Engineering Mechanics - Kinematics of Particle (KOP) - Discussion

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.

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.

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 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.

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.

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.

Here the car is not moving with uniform accleration upto ten seconds
area of the triangle=0.5*8*10=40m/sec(t=10)
now accleration is constant

So apply v=u+at
50=40+8*t
t=1.25

So total timet = 10+1.25 = 11.25

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.

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.

Dear, please tell me how it comes 40m/s. By which law and logic you calculate 40m/s.

FYI
1. V=u+at
3. v2=u2+2as
2. s=ut+1/2at2
4. s=1/2(v+u)t