# data[].attributes.body
A stone is thrown in a vertically upward direction with a velocity of 5m/s if the acceleration of the stone during its motion is 10m/s*s in the downward direction what will be the height attained by the stone and how much time will it take to reach there?
Ans) a =5
a=-10
V=0
By using
V=u+at
0=5+(-10)t
5=-10t
T=1/2=0.5
S=ut+ 1/2 at "2
S=5(1/2)+1/2(-10)×1/4
S=5/2.5/4).m
S=(10-5)/4 m
S=5/4m
-s=1.25m

# data[].attributes.body
A racing car has a uniform acceleration of 4m/s*s what distance will it covers in 10s after start
Ans) uniform acceleration a=4m/s
Initial velocity u=0m/s
Time+=10sec
Velocity v after 10 s =u+at=0m/s+4m/s square×10s
=40m/s
We shall calculate the distance covered by all possible method s
Using average velocity
Initial velocity u=0m/s
Average velocity=1/2(u+v)=1/2(0+40)=20m/s
Time=10s
Distance covered in 10s=AV.vel×time
=20m/s×10s =200m
Using the relations =ut+1/2 at 2
U=0m/s,a=4m/s square,t=10sec
S=0×10+1/2 4×10 square=200m
Using the relations (v square-u)-1/2×4(40square-0 square)
1/8(1600-0)
1600/8=200m

# data[].attributes.body
A trolley while going down an inclined plane has an acceleration of 2cm/s*s what will be it's velocity 3s after the start?
Ans) acceleration a=2cm/s"2=0.02m/s"2
To find?
Velocity after 3s=?
Solution:-
The trolley is on the inclined plane thereby the first equation of motion is to be applied to calculate velocity after 3s
We have:-
T=3sec
U=0
A=2cm/s square
V=0+(0.02×3)
=V=0.06m/s

# data[].attributes.body
2)A train travelling at a speed of 90km/ h, brakes are applied so as to produce a uniform acceleration of 0.5m/s.find how far the train will go before it is brought to rest ?
Ans) speed of train ,u=90km/h=90×5/18
=25m/s
Acceleration=a=0.5m/s square
Use formula=v=u+at
Finally train will be rest so final, velocity,v=0
0=25-0.5t
25=0.5t
•t=50 sec
Again use formula
S=ut+ 1/2at square
S is distance travelled before stop
S=25×50=1/2×0.5×50 square
1250=1/2×0.5×2500
1250-625=625m
Hence distance travelled=625m and time taken =50s

# data[].attributes.body
A bus starting from rest moves with a uniform acceleration of 0.1m/s*s for 2minutes find
a)the speed acquired
b)the distance travelled
Ans)given:-
Initial velocity of a bus (u)=0m/s
Acceleration gain by a bus,(a)=0.1ms-2
Time taken by the bus (t)=2minute
=120seconds
To find
a)the speed acquired
b)the distance travelled
Formula:-v=u+at
And
S=ut+ 1/2at square
Solution:-
The speed acquired
V=u+at
•v=0+0.1×120m/s
•v=12m/s
Hence the bus will acquire a speed of 12m/s after 2 months with the given acceleration
b)the distance travelled
S=it+1/2 at 2
=0×120+1/2×0.1×(120)
=1/2×0.1×14400m
=7.20m
Hence bus will travel a distance of 720m in the given time of minute