Question

A 1000 kg vehicle moving with a speed of 20 m/s is brought to rest in a distance of 50 metres:
(i) find the acceleration;
(ii) calculate the unbalanced force acting on the vehicle and
(iii) the actual force applied by the brakes may be slightly less than that calculated in (ii), give reason.


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Answer 1

Answer:

Vehical Mass = 1000 kg

Initial Velocity (that is speed) let it be u = 20 ms-1

Final Velocity let it be v= 0ms-1

Distance travelled = 50 m

let the acceleration be "a"

From equation => v2 = u2 + 2as

putting all value

(0m/s)2 = (20m/s)2 + 2*a*50 m

0m2/s2 = 400m2/s2 + 100a m

0m2/s2 - 400m2/s2 = 100a m     (Side changing)

-400m2/s2 = 100a meter

-400m2/s2 / 100 m = a          (side changing and cancelling m with m2)

-4m/s2 = a

a = -4 m/s2

This shows that the acceleration is in opposite direction that means it is retardation.

retardation = 4m/s2

Explanation:

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Answer 2

Answer:

Given,

Initial velocity(u)=20ms

−1

Final velocity(v)=0

Distance moved(S)=50m

We know,

v

2

=u

2

+2aS

0=20

2

+2a×50

−400=100a

a=

100−400

a=−4ms −2

Now,

F=ma

F=1000×−4

=−4000N