Question

A car starts from rest and accelerates uniformly for 20 seconds to a velocity of 72 km h^-1. It then runs at constant velocity and finally brought to rest in 200 m with a constant retardation. The total distance covered is 600 m. Find the acceleration, retardation and the total time taken

Answer 1

(i) Motion with uniform acceleration

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

∴v=u+at1

20=0+a×20 or a=1ms−2

distance travelled by car in this time (20 sec)

S1=ut+12at2=0+12×1×(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')×200

a'=-1 ms−2

Let t'= time for which the body comes to rest

∴ v=u+a't

0=20-1t'

∴ t'=20 sec

Answer 2

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Explanation:

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