Engineering Mechanics - Kinematics of Particle (KOP) - Discussion

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.

Acceleration * time = velocity.
So,in the above graph find the area covered by line up to uniform acceleration.

i.e., Area of triangle = 1/2 * 10 * 8 = 40 m/s. And then it follows constant acceleration and the required area(rectangle) is 10. So 8 * 1.25 = 10 m/s, so a total of (40 + 10)m/s is achieved in (10 + 1.25)secs.

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.

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.

The answer is incorrect correct answer is 6.25s.

Solution u = 0.

v = 50 m/s.
a = 8 m/s^2.

So just find t by v = u+at.
Which is 50 = 0+8t?

50 = 8t.
t = 50\8 = 6.25s.

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

How to calculate 40? which is v?

The area in the graph is velocity i.e. v=at, so the total area is given i.e.final velocity 50 m/s,

So 50= area of triangle + area of rectangle.

50= (0.5 x base x height) + (width x height).

50= (0.5 x 10 x 8) + [ (t-10) x 8].

Finally t=11.25 seconds.

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

That's very simple...

We know dat a= dv/dt
or we can say that v= integral(a dt)...............(1)
Now by graph we can see that a=(4/5)t upto the first 10 seconds
Now put v=50 in the eq.(1)

So after calculating t= 11.25 sec.