Acceleration in Real Life by MARIA CARMELA GEROMO

rosalymaeputian14 — 2021-02-25 13:52:27 UTC

An airplane begins by resting on the runway. For 53 × 10^-3 seconds the plane accelerates uniformly until it reaches an airspeed of 200 km/h and lifts off the runway.
a) Calculate the airplane acceleration.
b) Calculate the minimum runway length (ft) required for this takeoff.

GIVEN:
V= 200km/h
V_⁰= 0
t= 53 × 10^-3s

a)
V= (200km/h) (0.2778m/s / 1km/h) = 55.56m/s

V= V₀ + at
a= (V-Vo) / t
a= (55.56m/s - 0m/s) / 53 × 10^-3s
a= 55.56m/s / 53 × 10^-3s
a= 1.05 × 10^-3m/s²

b)
x= x₀ + v₀t + ½at²
x= 0 + 0 + ½ (1.05 × 10^-3m/s²) (53 × 10^-3s)²
x= ½ (1.05 × 10^-3m/s²) (2.809 × 10^-3s²)
x= ½ (2.94945 × 10^-6m)
x= 1.47 × 10^-6m

m -> ft
x= 1.47 × 10^-6m (3.28 × 10⁰ft / 1 × 10⁰m)
x= 4.82 × 10^-6ft

REAL LIFE ACCELERATION PROBLEM

saberonprincess08 — 2021-02-26 01:48:35 UTC

A stone is dropped into a deep well and is heard to hit the water 3.41 s after being dropped. Determine the depth of the well.

Given:
a = -9.8 m/s²
t = 3.41 s
v_i = 0 m/s
Find:
d = ??
Solution:

d = v_it + 0.5(at²)

d = (0 m/s)(3.41 s)+ 0.5(-9.8 m/s²)(3.41 s)²

d = 0 m+ 0.5(-9.8 m/s²)(11.63 s²)

d = -57.0 m

Therefore, the well has a depth of 57.0 m from the ground.

nielrelleve02 — 2021-02-26 13:51:09 UTC

LesterSatiada — 2021-02-27 01:19:23 UTC

REAL LIFE ACCELERATION PROBLEM
Alexa and Siri rent a room on a 4th floor at five storey apartment. They enjoy the beautiful view of the whole city through out the window. Alexa decided to take a selfie with her hand outside the window but suddenly the cellphone slipped on her hand and fell it to the ground. If the window is 12.5 m above the ground, determine the time required for the phone to reach the ground.
Given: d= -12.5m
a= -9.8m/s^2
Vi= 0m/s
Find: t
Solution: d= Vi t + 1/2at^2
d= 1/2at^2
t^2= (2)(d) ÷ a
t= √(2)(-12.5m)÷(-9.8m/s^2)
t= √(-25m)÷(-9.8m/s^2)
t=√2.55
t= 1.6 s
Therefore, the cellphone takes 1.6 seconds before it reaches the ground.

boyetmora68 — 2021-02-27 04:14:48 UTC

kyryllbebis9 — 2021-02-27 06:04:05 UTC

Real life problem on uniform accelerated motion and freefall

A group of friends went out of town to celebrate their friendship anniversary. They tried the cliff jumping and recorded it. It took one to reach the seawater after 1.3 s. What is the velocity before he reaches the seawater? How high is the cliff? Convert distance to ft.

Given:
t = 1.3 s
vi = 0 m/s
g = 9.8 m/s^2

Find:
a.) Vf
Vf = Vi + gt
= 0 m/s + 9.8 m/s^2 (1.3 s)
= 12.7 m/s

b.) d
d = Vit+ gt^2 / 2
= 0 m/s (1.3 s) + 9.8 m/s^2 (1.3s)^2 / 2
= 9.8 m/s^2 (1.69 s ^2) / 2
= 16.56 / 2
= 8.3 m

c.) m to ft
1 m = 3.28 ft
8.3 m * 3.28 = 27.2 ft

REAL LIFE ACCELERATION

cjyhoann — 2021-02-27 06:42:56 UTC

Miguel's crop duster plane must reach a speed of 120mi/h to take off, if Miguel starts at a rolling speed of 5.00mi/h and accelerates at 7.55ft/s, how long must his take off field be?

Given:
Vf = 120mi/h
Vi = 5mi/h
A = 7.55ft/s^2
Find:
D =?
Equation:
Vf^2 = Vi^2 + 2ad

Fill in equation:
(120mi/h)^2 = (5mi/h)^2 - 2(7.55ft/2^2)(d)

Work/Answer:
14400mi^2/h^2 = 25mi^2/h^2 +15.1ft/s^2(d)
14400mi^2/h^2 - 25mi^2/h^2 +15.1ft/s^2(d)
13375mi^2/h^2 = 15.1ft/s^2(d)
13375mi^2/h^2 = D
_______________
15.1ft/s^2
952mi^2/h^2 × s^2/ft =d

D = 0.39mi or 2.0 × 10^3ft

mariancornelio028 — 2021-02-27 07:49:12 UTC

Real Life Acceleration Problem

magistradoac — 2021-02-27 08:00:01 UTC

Ma. Kyryll wants to try cliff jumping and she decided to go on a popular cliff diving spot in Batason, Buruanga, Aklan. She jumps from cliff into water below. The height of the cliff is 8m, how long will it take Kyryll to hit the water?

Given: d=8m
Vi= 0
a=g= 9.81 m/s²Find
Find: t=?

Solution:
d= Vit+1/2at²
d= 1/2at²
t= √2d/a
t= √2(8m)/ 9.81m/s²
t= √16m/ 9.81m/s²
t= √1.63
t= 1.28s

bragaismarklester02 — 2021-02-27 08:08:43 UTC

REAL-LIFE ACCELERATION PROBLEM

Angelo is riding a bicycle with a constant acceleration of +4m/s². If the initial velocity of Angelo's bicycle is 20m/s, (a) what is the displacement of hks bicycle after 15 seconds? (b) What was his final velocity in riding a bicycle?

GIVEN:
Vo = 20m/s
t = 15s
a = +4m/s²
FIND:
d = ?
V = ?

A.
d = Vot + ½at²
= (20m/s)(15s) + ½(+4m/s²)(15s)²
= 300m + 450m
d = 750m

B.
V = Vo +at
= 20m/s + (+4m/s²)(15s)
= 20m/s + 60m/s
V = 80m/s

reodiqueluther12 — 2021-02-27 08:52:18 UTC

REAL LIFE ACCELERATION PROBLEM

cashazel18 — 2021-02-27 09:00:58 UTC

Vy, Prince, and Cess decided to ride a Ferris wheel. Their seat is going up high, and they can now feel the tension as the Ferris wheel keeps orbiting. While on top, Vy's slipper fell off. It took 10s to reach the ground. Determine how high their seat is from the ground.

Given:
t = 10s
a = -9.8 m/s²
Vi = 0m/s

Find: d

Solution:
d = Vi + ½ at²
= (0m/s)(10s) + ½ (-9.8 m/s²)(10s)²
= 0 + ½ (-9.8 m/s²)(100s²)
d = - 490 m

Therefore, their seat was 490m above ground when the slipper fell off.

REAL LIFE ACCELERATION PROBLEM

ybescurel06 — 2021-02-27 10:11:43 UTC

Yuro rolled down his pencil and fell, with a velocity of 5 m/s straight forward on the desk. If the desk height is 0.76 m below the desk, how long will it be in the air before it meets the floor?

Find the final velocity
Find the time

Given: Vo=5 m/s

Vf=

d=0.76 m

a=-9.8m/s^2

Vf^2=Vo^2 + 2ad

= (5m/s)^2 + 2(-9.8 m/s^2) (0.76m)

= 25 m/s + (-14.896 m/s^2)

Vf = 10.104 m/s

g= Vf -Vo/ t

t = Vf -Vo/ g

t= 10.104m/s - 5 m/s / -9.8 m/s^2

t= 0.52 s

hernandezaerieljoimaika — 2021-02-27 10:34:49 UTC

If a basketball player has a vertical leap of 1.29 m, then what is his takeoff speed and his hang time (total time to move upwards to the peak and then return to the ground)?

Given:
a = -9.8 m/s²
vf = 0 m/s
d = 1.29 m

Find:
vi = ??
t = ??

vf² = vi² + 2ad
(0 m/s)² = vi² + 2(-9.8 m/s²)(1.29 m)
0 m²/s² = vi² - 25.28 m²/s²
25.28 m²/s² = vi²

vi = 5.03 m/s

To find hang time, find the time to the peak and then double it.

vf = vi + at
0 m/s = 5.03 m/s + (-9.8 m/s²)tup
-5.03 m/s = (-9.8 m/s²)tup
(-5.03 m/s)/(-9.8 m/s²) = tup

tup = 0.513 s
hang time = 1.03 s

francinerastrullo0912 — 2021-02-27 11:35:15 UTC

A frog is capable of jumping to a height of 3.5 m. Determine the takeoff speed of the frog.

Given:

a = -9.8 m/s²

vf = 0 m/s

d = 2.62 m

Find:

vi = ??

vf²= vi² + 2*a*d

(0 m/s)²= vi² + 2*(-9.8 m/s2)*(3.5 m)

0 m²/s² = vi² - 68.6 m²/s²

68.6 m²/s² = vi²

vi = 8.28m/s

REAL LIFE ACCELERATION PROBLEM

villarshienamae2 — 2021-02-27 14:46:05 UTC

A window cleaner dropped his spray bottle from the window of the 20th floor of a building and it took 5 seconds before it hit the ground. Calculate the spray bottle's velocity and the height of the 20th floor window.

Given:
Vi = 0 t = 5s
g= 9.8m/s^2

Find:
a. V=?
b. d=?

a. V = Vo + gt = 0 + (9.8m/s^2)(5s) = 49 m/s

b. d = Vot + gt^2/2 = 0 + (9.8m/s^2)(5s)^2/2 = 122.5m

ericanicolenorte6 — 2021-02-27 15:00:44 UTC

A piece of coconut fruit falls freely from a coconut tree which takes 2.6567 seconds to reach the ground. Calculate the velocity before the coconut fruit reach the ground and find how tall is the coconut tree in ft.

Given:
g = 9.8m/s²
t = 2.6567s
V0 = 0m/s

Find:
V = ?
d = ?

Solution:
V = V0 + gt
V = 0m/s + (9.8m/s²) (2.6567s)
V = 26.04m/s²

d = V0t + gt² / 2
d = 0 + (9.8m/s²) (2.6567s)² / 2
d = (9.8 m/s²) (7.05805489s²) / 2
d = 69.168937922 m / 2
d = 34.58m

Convert m to ft:
34.58m × 3.28084ft / 1m = 113.45 ft

delacruzlloydgino — 2021-02-27 16:21:44 UTC

aerlzenjulessabando — 2021-02-28 01:48:04 UTC

The coconut fell from the coconut tree and it took 1.56 seconds to hit the ground. Neglecting the air resistance. How high the coconut to ground? What is the velocity of the coconut before it hit the ground? Given: t= 1.56 seconds
a= -9.8m/s
V0 = 0
Y0 = 0
Y = ? V= ?
Y= Y0 + V0 + ½ at2
Y= 0+ 0(1) + ½ (-9.8m/s2)(1.56s)2
Y= 11.92m, therefore 11.92m is the high of the coconut to the ground before it fall.
V= V0 + at V= 0(1)+ (-9.8m/s2)(1.56s)
V= 15.29m/s is the coconut velocity before hitting the ground.

2021-02-28 05:34:34 UTC

My 7 year old brother spotted his cat at Manang Esther's roof which is 10.0m above him. A firewood have been launched vertically into the air with a velocity of 5.6m/s neglecting air resistance for he knows that his cat will reflexively drop from the roof to the floor immediately after he released the firewood:
a)how high does the firewood go? b.find the impact velocity of a cat when it hits the ground surface.

Solution:
Vf²=Vi²± 2ad

Given
a.Vi= 5.6 m/s
a= -9.81 m/s²
∆y = 10.0 m
d= ?
Vf²=Vi²± 2ad
O=Vi² - 2ad
2ad= Vi²
d= Vi²/2(a)

d= Vi²/2(a)
d=(5.6m/s)² / 2(-9.81 m/s²)
d=31.36 m/s / 19.62 m/s²
d=1.61 m + 10.0 m
d=-11.61 m

Given
b.Vi= 0m/s
a= -9.8 m/s²
d=11.61 m
Vi= ?
Vf²= Vi²± 2 ad
Vf²=0m/s+ 2(-9.81m/s²)(-11.61m)

Vf=√2(-9.81m/s²)(-11.61 m)
Vf=√227.78 m/s
Vf=√227.78 m/s

2021-02-28 05:35:09 UTC

REAL-LIFE ACCELERATED MOTION

angelinajolietibus — 2021-02-28 05:41:03 UTC

Angelina planned to go on her crew's skate park/spot to train to improve her skateboarding skills in the preparation for the Game of Skate Competition where she and her crew will participate. She used her skateboard as her means of transportation to go on their spot, since it is only nearby. She went out on their house and started skating. She is riding her skateboard with constant acceleration of +3m/s^2. If the initial velocity of Angelina's skateboard is 10m/s(a), what is the displacement of her skateboard after 25 seconds? What was her final velocity in riding her skateboard?

Given: Vi = 10m/s
t = 25s
a = +3m/s^2
Find: d=?
Vf=?

A. d=Vo + 1/2 at^2
=(10m/s)(25s) + 1/2(+3m/s^2)(25s)^2
=250m + 937.5m
=1187.5m
B. Vf=Vo + at
=10m/s + (+3m/s^2)(25s)
=10m/s + 75m/s
Vf=85m/s

zoilamarie111 — 2021-02-28 05:55:33 UTC

ZOILA MARIE S. RUBIO
My 7 year old brother spotted his cat at Manang Esther's roof which is 10.0m above him. A firewood have been launched vertically into the air with a velocity of 5.6m/s (neglecting air resistance) for he knows that his cat will reflexively drop from the roof to the floor immediately after he released the firewood:
a)how high does the firewood go? b.find the impact velocity of a cat when it hits the ground surface.

Solution:
Vf²=Vi²± 2ad

Given
a.Vi= 5.6 m/s
a= -9.81 m/s²
∆y = 10.0 m
d= ?
Vf²=Vi²± 2ad
O=Vi² - 2ad
2ad= Vi²
d= Vi²/2(a)

d= Vi²/2(a)
d=(5.6m/s)² / 2(-9.81 m/s²)
d=31.36 m/s / 19.62 m/s²
d=1.61 m + 10.0 m
d=-11.61 m

Given
b.Vi= 0m/s
a= -9.8 m/s²
d=11.61 m
Vi= ?
Vf²= Vi²± 2 ad
Vf²=0m/s+ 2(-9.81m/s²)(-11.61m)

Vf=√2(-9.81m/s²)(-11.61 m)
Vf=√227.78 m/s
Vf=√227.78 m/s

nudanicole1 — 2021-02-28 06:51:13 UTC

I hit the tennis ball crosscourt so that my opponent will move with an initial velocity of +7.93 m/s straight upward. If the tennis ball starts from 170cm above the floor, how long will it be in the air before it strikes the ground?

Given:
V0 = +7.93m/s
∆Y = -1.7cm
g = -9.8 m/s²
V = ?
t = ?

Solution:
Covert first cm to m
(-170cm) (0.01m / 1cm) = -1.7m

V² = V0² + 2ad
V² = (7.93m/s)² + 2(-9.8m/s²) (-1.7m)
V² = 62.8849 m/s² + (-19.6 m/s²) (-1.7m)
√V² = √96.2049 m/s²
V = 9.81 m/s

g = V - V0 / t
t = V - V0 / g
t = -9.81 m/s - 7.93m/s / -9.8m/s²
t = 1.81s

mcdave7154 — 2021-02-28 09:26:12 UTC

jescelsaberola — 2021-02-28 11:25:02 UTC

Real Life Acceleration Problem

ACCELERATION IN REAL LIFE PROBLEM

2021-02-28 12:08:56 UTC

Pro tip:

Shallow-water waves move at a speed, c, that is dependent upon the water depth and is given by the formula:

where g is the acceleration due to gravity (= 9.8 m/s2) and H is the depth of water.

Problem:

Mount Unzen, an active volcanic group of several overlapping stratovolcanoes in Japan, erupted. After an initial eruption, a large earthquake triggered a landslide from the Mayuyama peak, a 4,000-year-old lava dome rising above the city of Shimabara. The massive landslide swept through the city and eventually reached the Ariake Sea where it set off a tsunami with a speed of 161.7m/s and a constant acceleration of + 0.27 m/s², from rest, at the deep-waters of the open sea. When will the tsunami reach the shallower waters near the coast?

Given: at constant acceleration:
V₀= 0m/s

a= +0.27m/s² V= 161.7m/s
t=? in minutes

Solution:

V=V₀+at
t=V-V₀/a

t=161.7m/s-0m/s/0.27m/s²

t=161.7m/s/0.27m/s²

t=599s or approximately 10mn

Therefore, the tsunami will reach the shallower waters near the coast in about 10 minutes.

ACCELERATION PROBLEM IN REAL LIFE

gensilerio3 — 2021-02-28 12:21:20 UTC

Problem:
Mount Unzen, an active volcanic group of several overlapping stratovolcanoes in Japan, erupted. After an initial eruption, a large earthquake triggered a landslide from the Mayuyama peak, a 4,000-year-old lava dome rising above the city of Shimabara. The massive landslide swept through the city and eventually reached the Ariake Sea, with a depth of 165m, where it set off a tsunami with a constant acceleration of + 0.27 m/s2 , from rest, that propagates at the deep-waters of the open sea. When will the tsunami reach the shallower waters near the coast?

Given: at constant acceleration:
V0= 0m/s
a= +0.27m/s2 V= ?
t=? in minutes
Solution:
a. C=V
V=√gh
V=√9.8m/s² x 165m
V=40.21m/s

b. V=V0+at
t=V-V0/a
t=40.21m/s-0m/s/0.27m/s2
t=40.21m/s/0.27m/s2
t=149s or approximately 2.5 mn

Therefore, the tsunami will reach the shallower waters near the coast in about 2.5minutes.

Real Life Acceleration Problem

mingsebastian19 — 2021-02-28 12:41:30 UTC

It was New Year’s Eve when Junior, a Policeman decided to shoot a bullet from his riffle straight upward from ground level. The bullet left the barrel at an initial speed of 630 m/s. Determine (a) the highest point it reaches, (b) the time it took to reach the highest point and (c) the speed before landing back to the ground. Neglecting Air Resistance.

Solution

Given:

Since at the highest point the bullet stops, V_f = 0

V_i = 630 m/s

Y =?

T =?

V_f² – V_i² = 2 g y

0² – 630 m/s = 2 (9.8) y

396900/19.6 = 19.6/19.6 y

Y = 20250m

T = (V_f -V_i) g

T = (0 – 630) 9.8

T = 64s

V_f = g t

V_f = (-9.8) 64

V_f = -627.2 m/s (negative sign to indicate that the bullet is moving downward)

brozoearlgerard — 2021-02-28 15:00:50 UTC

Rens R. Recio

snersoroicer — 2021-02-28 18:06:39 UTC

REAL LIFE ACCELERATION PROBLEM

ajerochristine04 — 2021-02-28 18:39:02 UTC

Real Life Acceleration Problem

bondaddanielyn — 2021-03-01 00:23:53 UTC

REAL LIFE ACCELERATION PROBLEM

uyclarsv09 — 2021-03-01 07:40:04 UTC

REAL LIFE ACCELERATION

2021-03-01 13:54:28 UTC

The observation deck of tall skyscraper 370 m above the street. Determine the time required for a penny to free fall from the deck to the street below.

Given:

v_i = 0 m/s
d = -370 m
a = -9.8 m/s^{2
✔}Find:t = ??

SOLUTION:
d = v_i*t + 0.5*a*t²-370 m = (0 m/s)*(t)+ 0.5*(-9.8 m/s²)*(t)²-370 m = 0+ (-4.9 m/s²)*(t)²(-370 m)/(-4.9 m/s²) = t²75.5 s² = t²t = 8.69 s

REAL LIFE ACCELERATION

sabbyreyes09 — 2021-03-01 14:00:35 UTC

The observation deck of tall skyscraper 370 m above the street. Determine the time required for a penny to free fall from the deck to the street below.

GIVEN:

v_i = 0 m/s
d = -370 m
a = -9.8 m/s²

Find: t = ??

Solution:
d = v_i*t + 0.5*a*t²-370 m = (0 m/s)*(t)+ 0.5*(-9.8 m/s²)*(t)²-370 m = 0+ (-4.9 m/s²)*(t)²(-370 m)/(-4.9 m/s²) = t²75.5 s² = t²t = 8.69 s

Will the drunk man stay alive? Caaya levi III

2021-03-01 14:48:50 UTC

PNR train moves at a velocity of 50 meters/minute. After reaching the station in Basud, Polangui Albay it begins accelerating at constant acceleration rate of 120 meters/minute2. There's a drunk person in batangay sugcad that is 777 meters away from the station, a physics student is worried if that man will safely cross the rails that he needs to walk a distance of 5 meters, he is also constantly accelerating at the rate of 5 meters per minute. Will the student save his life?
We need to calculate for the time that train will reach man.
But we do not have the final velocity so we need to compute final velocity.
1. A. Given Vi = 50 m/min
a = 120 m/min^2
d = 777 meters.
V^2 = Vi^2 + 2ad
= (50)^2 + 2 (120)(777)
= 2500 + 93, 240
V^2 = 95740
Square root of 95 740
= 309 m/ min this will be the final velocity
Now we compute for the time
B. V = V0 + at
t = (V - V0)/a
= (309 - 50)/120
= 2.16 minutes will be the time the the train will reach at 777 meters away from basud.
Now we calculate thw drunk man's safety in determiningif he crossed the rail faster before the train kills him.
We also need to computefor the final velocity
2. Given
a = 30 meters/ min
d = 5 meters
Vo = 0 m/s

V^2 = V0 ^2 + 2ad
= 0 + 2(30)(5)
= 300
Square root of 300 = 17 meter/min

Compute for time
t = (V-V0)/a
=( 17-0)/5
3.4 minutes.
2.6 < 3.4
The train will arrive faster before the man crosses the rails, he might be in danger so advisimg him to stop will be the student's approach on the situation and also his conscience.

crissymaricernal — 2021-03-01 14:52:44 UTC

elaurzadiana — 2021-03-01 15:24:21 UTC

REAL LIFE ACCELERATION PROBLEM

tagumKez — 2021-03-02 03:39:59 UTC

REAL LIFE ACCELERATION PROBLEM
Quezzalyn Ayn A. Tagum
12 STEM EUCLID

A toy car starts from rest and accelerates at a constant rate of 7.56 cm/s² for 2.6 seconds then breaks at 1.5 seconds and experiences an acceleration of -8.07 m/s². A) How fast is the car going at the end of the breaking period? B)How far has it moved?

Given:

V1i = 0
t1 = 2.5 s
a1 = 7.56 cm/s²
V2i = ?
t2 = 1.5 s
a2 = -8.07 cm/s²

A)
V1f = Vi + at
V1f= 0+(7.56cm/s²)(2.5s)
= 14.56cm/s
V1f = V2i
V2f = V2i + a2t2
= 14.56 cm/s + (-8.07cm/s²)(1.5s)
= 2.46 cm/s

A) 2.46 cm/s is the velocity of the car at the end of the breaking period

B)

Δx1 = ½(V1f + V1i) t1
Δx1 =½(14.56cm/s + 0) (2.6s)
= 18.93 cm

Δx2 = V2i Δt2 +½ a2(Δt2)²
= 14.56m/s (1.5s) +½(-8.07cm/s²)(1.5s)
= 12.76 cm
Δxf = Δx1 + Δx2
= 18.93 cm + 12.76 cm
= 31.69 cm
B) 31.69 cm is the distance of how far the toy car moved.

nacariokyla09 — 2021-03-02 05:43:15 UTC

In a cheerleading stance, Lyla is lifted 1.8m from the ground. If she was thrown upward and catched after 1.4s, how high did she reached above the ground?

GIVEN:
t = 1.4 s
g = 9.8m/s²
X = 1.8 m
d = ?

SOLUTION:
d = Vot + gt²
d = (0m/s)(1.4s) + ½(9.8m/s²)(1.4)²
d = 9.6 m

9.6m + 1.8m = 11.4m

Lyla reached 11.4m above the ground.

2021-03-04 00:53:08 UTC

Sophia S. Retis
12 Stem Euclid

Paolo Simeon O. Satuito

simeonpaolo2002 — 2021-03-04 04:05:45 UTC

12-STEM Euclid

kennethnavarro899 — 2021-03-04 06:40:35 UTC

Angeline R. Bodino

2021-03-04 08:57:02 UTC

An Aircraft starts from and accelerates at 1.5 m/s² until it takes off at the end of the runway.What was the final speed at the end of the 1.8 km runway?

V2²=V1²+ 2ad a=1.5 m/s²
=0+(2)(1.5)(1800) V1= 0
=5400 d=1.8 km = 1800 m
V2=√5400 V2=?
V2=73.48469
:the final speed at the end of the runway is 73m/s

ACCELERATION IN REAL LIFE

camoralseantara09 — 2021-03-04 12:08:14 UTC

Olive and Paul are on their way to a beach resort using the family car as their transportation. Their car starts from rest and accelerates uniformly over a time of 5.21 seconds for a distance of 110 m. Determine the acceleration of Olive and Paul’s car.

Given:
d = 110 m
t = 5.21 s
v_i = 0 m/s

Find:
a = ?

SOLUTION:
d = v_i*t + 0.5*a*t²

110 m = (0 m/s)*(5.21 s)+ 0.5*(a)*(5.21 s)²

110 m = (13.57 s2)*a

a = (110 m)/(13.57 s²)

a = 8.10 m/ s²

Therefore, the the acceleration of Olive and Paul's car is 8.10 m/ s²

rhnrodelo — 2021-03-04 12:36:22 UTC

2021-03-04 23:54:00 UTC

Jann Nathaniel S. Rellon

REAL LIFE ACCELERATION PROBLEM A bowling ball falls freely (near the surface of the earth) from rest. How far does it fall in 5 seconds, and how fast will it be going at the time? Given: Vi=0 t=5s Solution v=go + gt= 0 + (9.8m/s^2)(5s)=49 d=vot+gt^2/2= 0+(9.8/s^2)(5s)/2=24.5 XREAL LIFE ACCELERATION PROBLEM A bowling ball falls freely (near the surface of the earth) from rest. How far does it fall in 5 seconds, and how fast will it be going at the time? Given: Vi=0 t=5s Solution v=go + gt= 0 + (9.8m/s^2)(5s)=49 d=vot+gt^2/2= 0+(9.8/s^2)(5s)/2=24.5 REAL LIFE ACCELERATION PROBLEM A bowling ball falls freely (near the surface of the earth) from rest. How far does it fall in 5 seconds, and how fast will it be going at the time? Given: Vi=0 t=5s Solution v=go + gt= 0 + (9.8m/s^2)(5s)=49 d=vot+gt^2/2= 0+(9.8/s^2)(5s)/2=24.5 REAL LIFE ACCELERATION PROBLEM

2021-03-07 06:39:48 UTC

A bowling ball falls freely (near the surface of the earth) from rest. How far does it fall in 5 seconds and how fast it be going at that time?

Given
v=Vi+Vo
0+(9.8m/s^2)(5s)=49m/s

d=Vot+gt^2/2
0+(9.8m/s^2)(5s)=78.4m