Engineering Mechanics - Kinematics of Particle (KOP) - Discussion

Initially the is at rest, we consider u=40m/s, and v=50m/s then

v=u+at
50=40+8*t
t=1.25sec,

This is the time required to travel from 40m/s velocity to 50m/s velocity. then the time required to travel the car from rest position is 10+1.25=11.25 sec.

Actually we want calculate and differentiating function, like ankith what he was explain.

Aishwarya Katale : Actually 40m/s its area under the curve. :-).

No, 11.25 is correct.

The formula v = u + at is applicable only in the case of constant acceleration.

Since up to the first 10 seconds the car is moving wth increasing acceleration area under the triangle gives final velocity at t =10 sec since initial velocity is 0.

11.25 secs are the correct answer.

Equations of motion are:

v = u + at;.

s = ut + 0. 5 at^2 etc are valid when accleration is constant.

In the question, the acceleration varies with timeup to 10 sec and then it is constant so, use
v = u + at after 10 secs.

And for before 10 secs use an area under a - t graph is the velocity similarly area under v - t graph gives displacement.

You will understand this concept when you pass through basics of calculus.

1. We consider the given acceleration and time (i.e a=8 ms-2 t=10s) then we find velocity for that next we pick any one choices given and add or sub for making it equal to 10 find another velocity.