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Question

Solve the following:

A) An engine is moving with a velocity 44 m/s. After applying the brakes, it stops after covering a distance of 121 m. Calculate retardation and time taken by the engine to stop.

B) A body starts from rest and moves with a constant acceleration. It travels a distance S1 in first 10s, and a distanceS2 in next 10s. Find the relation between S1 and S2. [5 MARKS]

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Solution

Part A : 2 Marks
Part B : 3 Marks

A)
Here,
u=44m/s, S=121m, v=0
Using, v2=u2+2aS
0=442+(2×121×a)
a=8m/s2

Also,
v=u+at
0=448t
t=5.5s


B)
Given :
u=0, t1=10s
Let the acceleration be a.
Distance travelled in first 10 seconds, is given by
S1=ut+at22
= 0+a×10×102
S1=50a ------- (1)

To calculate the distance travelled in next 10s, we first calculate distance travelled in 20 s and then subtract distance travelled in first 10 s.
S=ut+at22
= 0+a×20×202
= 200a

Distance travelled in the next 10s interval,
S2=SS1=200a50a
S2=150a

Using equation 1,
S2=3S1


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