Question

a particle starts moving with velocity 10 m/s in a straight line under an acceleration varying linearly with time. its velocity time graph is as shown in figure. its velocity is maximum at t=3 sec. the time when the particle stops is (tan 37=3/4) 1. 6 2. 8 3. 7 4. 5

Answer 1

We know that the acceleration is given by the velocity time graph. So from the above curve we can see that the acceleration (a) at t=0, is tan(37)=34, and acceleration at t=3 is tan0=0. Also acceleration when the particle stops is a=tan(90+45)=tan(135)=−1
Since the acceleration is varying linearly with time suppose the equation of acceleration is given by
a=mt+c
Since at t=0, a=34, we have, c=34
and at t=3, a=0 gives m=−14
So when a=−1, we have
−1=−14t+34⇒t=1+3414=7
So velocity will be zero at 7 s. Hence answer 3. is correct answer.

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Answer 2

Answer:

The correct answer is option 3 ⇒ 7. The velocity will be zero at 7s.

Explanation:

We are aware that the velocity-time graph provides the acceleration.

The acceleration 'a' at t = 0, is tan(37) = 34

Acceleration at t = 3 is tan0 = 0

Acceleration when the particle stops a = tan(90+45)

= tan(135)

= −1

Assume that the equation for acceleration is provided by a = mt + c because the acceleration varies linearly with time.

At t = 0, a = 34, and we have, c = 34

At t = 3, a = 0 gives m = −14

When a = −1, we will have:

−1 = −14t + 34

t = 1 + 3414

= 7s

Therefore, at 7s, velocity will be zero.

In essence, velocity is a vector quantity. It is the speed at which distance changes. It is the displacement change rate. A moving item can never move at a negative speed. A moving item can have zero velocity.