Question

On a straight road, a car starts from rest and accelerates at uniform rate of 6 m/s² for some time, then moves with constant speed for some time and retards the same uniform rate and comes to rest. Total time for the joumey is 24 s and average speed for joumey is 20 m s How long does the car move with constant speed?

Answer 1

let the time for inital uniform acceleration be t.
therefore time for final uniform a
cceleration will also be t
u.acceleration(a)=6
inital velocity(u)=0 (body was at rest)
therefore time for constant speed =24-2t
velocity on start of conrtant speed=0+2as
=sq.root(2*6*3t^2)(from 2nd eq. of motion)
=sq.root(36*t^2 )
=6t

therefore distance covered by constant speed=
vt
=(24-2t)6t

distance covered in acceleration
=1/2*at^2
=3t^2

distance covered in deceleration
=vt+1/2*at^2
=vt-3t^2

average speed=total distance/total time
20=(3t^2+vt+vt-3t^2)/(t+t+24-2t)
20=2vt/24-2t
20(24-2t)=2*(24-2t)*6t
20=2*6t
t=20/12=5/3

therefore time for uniform speed=24-2*5/3
=24-10/3
=62/3=20.66 s

Answer 2

Answer:

See the pic fr solution

Explanation:

Total distance = distance travelled during acceleration, deceleration,at constant speed altogether

Total time 24 sec