Maths-
General
Easy
Question
The displacement of a particle in time ‘t’ is given by S = t3 – t2 – 8t – 18. The acceleration of the particle when its velocity vanishes is
- 10
- 10 units /sec2
- 5 units/sec2
- 20 units/sec2
Hint:
We are given the equation of displacement of a particle. We have to find the value of its acceleration at the point where velocity vanishes i.e. becomes zero. Acceleration is second order derivative of the equation and velocity is first order derivative of displacement.
The correct answer is: 10 units /sec2
The given expression is s = t3 - t2 - 8t - 18.
To get the velocity we will differentiate the above equation w.r.t "t" . And to find the point at which it's vanishing, we will equate the derivate with zero.


The point at which velocity is vanishing is t = 2
Now, we will find the equation for acceleration by taking the second order derivative of displacement.

We will substitute the value of t = 2 to find the acceleration at required point.
a = 6(2) - 2
= 12 - 2
= 10 units/sec2
So, the acceleration at the point where velocity vanishes is 10units/sec2
For such questions, we should know relation between displacement, velocity and time. We should also know how to find first and second order derivate.
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