Question

The displacement of a particle is represented by the following equation: \[s = 3{t^3} + 7{t^2} + 5t + 8\] where s is in meters and t in seconds. The acceleration of the particle at t=1 s is:-
A. \[14m/{s^2}\]
B. \[18m/{s^2}\]
C. \[32m/{s^2}\]
D. zero

Answer 1

Hint:Velocity can be defined as the rate of change of displacement of a particle. It can also be defined as the time derivative of displacement. So we differentiate the given displacement then we get the velocity of the particle. Similarly, acceleration can be defined as the rate of change in the velocity of a particle. It can also be defined as the time derivative of velocity. Hence again differentiating the velocity gives acceleration.

Formula used:
Velocity of a particle is given as:
\[velocity = \dfrac{\text{change in displacement}}{\text{time taken}}\]
\[v = \dfrac{{ds}}{{dt}}\]
Where s is displacement and t is the time taken.
Acceleration of a particle is given as:
\[acceleration =\dfrac{\text{change in velocity}}{\text{time taken}}\]
\[a = \dfrac{{dv}}{{dt}}\]
Where v is velocity and t is the time taken.

Complete step by step solution:
Given Displacement of a particle, \[s = 3{t^3} + 7{t^2} + 5t + 8\]
As we know velocity of the particle, \[v = \dfrac{{ds}}{{dt}}\]
By using the given displacement value,
\[v = \dfrac{d}{{dt}}(3{t^3} + 7{t^2} + 5t + 8)\]
\[\Rightarrow v = 9{t^2} + 14t + 5\]
As we know acceleration of the particle,
\[a = \dfrac{{dv}}{{dt}}\]
By using the above velocity value,
\[a = 18t + 14\]
Now put t= 1s, we get
\[a = 18 \times 1 + 14\]
\[\therefore a = 32\,m/{s^2}\]
Therefore, the acceleration of the particle at t=1 s is \[32\,m/{s^2}\].

Hence option C is the correct answer

Note: Acceleration is defined as the rate of change of velocity of a particle. A particle’s acceleration is the final result of all the forces being applied to the body, as defined by Newton’s second law. Acceleration is a vector quantity. In general, acceleration is defined as speeding up. Also, we say that it is the rate at which the velocity of an object changes. It does not depend if it is speeding up or down.