Question

a car starts from rest and accelerates uniformly for 20 seconds to velocity 72 kilometre per hour then once at constant velocity and finally comes to rest in 200 metres with constant acceleration if total distance is 600 find the acceleration and retardation total time taken for the journey

Answer 1

Answer:

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Explanation:

(i) Motion with uniform acceleration

Here , u=0, t1=20 sec, v=72×518=20ms-1

..v=utat1

20=0+ax20 or a=1ms-2

distance travelled by car in this time (20 sec)

S1=ut+12at2=0+12x1x(20)2=200m

(ii) Motion with uniform velocity. As given, total distance = 600 m we have calculated S1 = 200 m (with uniform acc.)

and S2 = 200 m (with retardation)

.: Net distance for which body moves with uniform velocity

S=600-S1-S2=600-200-200=200m

... Time taken t=distanceuniform velociy

=20020=10sec

:. Total time of journey, t=(20+10+20)sec

t=sec

Average velocity=Total displacementTotal Time=60050

=12 m/s

(iii) for this motion, initial velocity.

u=20ms-1 and

Final velocity v=0,S2=200m

Acceleration a'=?

Using v2-u2=2a'S2

(20)2=2(a')x20o

a'=-1 ms-2

Let t's time for which the body comes to

rest

... v=u+a't

O=20-1t'

.. t'=20 sec