Question

A car starts moving along a line first with an acceleration a =5m.s starting from rest then uniformly and finally decelerating at the same rate comes to rest in the total time of 25seconds t ,then average velocity during time is equal to v = 72kmph how long does the particle move uniformly?​

Answer 1

Answer:

➡️Let the initial velocityu=0

The time taken and the distance traveled during accelerated motion and retarding motion will be same as the acceleration is same

For accelerated motion

v=u+at=5t

And for retardation

0=v−5t

v=5t

As the velocity v will become u for retardation

Total distance traveled

= area of v−t

= areaI+ area II+ area II

$2( \frac{1}{2} 5t {}^{2} ) + 5tx = distance \: traveled \: during \: motion$

x=time during which particle remains in uniform motion

5t² +5tx=speedx time

5t²+5tx=20×25

t²+tx=100→1

Also the total time is given 25seconds

Thus 2t+x=25

Thus x=25−2t→2

Substituting 2 in 1

t²+t(25−2t)=100

t² −25t+100=0

t=5 or t=20

Thus the time will be t=5sec as t=20s is not possible as it will exceed the given time

The particle remains in motion for 5 seconds

Refer the attachement for diagram ⬆️⬆️⬆️⬆️

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