Question

The velocity of the body is given by $v=6-0.02t$,where t is the time taken. The body is undergoing:-
A. Uniform retardation of 0.02ms-2
B. Uniform acceleration of 0.02ms-2
C. Uniform retardation of -0.02ms-2
D. Uniform acceleration of - 0.02ms-2

Answer 1

When displacement ; velocity; acceleration equations are given in time function then for calculating the displacement , velocity , acceleration at particular time we have to use calculus method (differentiation and integration approach)for solving these equations. For solving this equation we cannot use three equations of motion.

Complete step-by-step answer:
Since acceleration of a body is defined as rate of change in velocity with respect to time. When acceleration is to be calculated from time based equation then calculus method is used while when time equation of velocity is not given then it is calculated by three equation of motion.

When displacement equation is given in time function then only velocity can be calculated by using differentiation equation .
$v=\dfrac{dx}{dt}$.
When velocity equation is given in time function then acceleration is calculated by using differentiation equation
$a=\dfrac{dv}{dt}$.(Equation 1)
Then Displacement can also be calculated from velocity time equation by using this relation.
$x=\int{vdt}$.
When acceleration equation is given in time equation then only velocity can be calculated by using this relation.
$v=\int{adt}$
Solving of all differentiation equations and integration equations can be done with respect to time.
According to given Question,
Velocity of body is given by the equation as $v=6-0.02t$.
So for calculating acceleration/retardation we have to use the above velocity equation for solving it by differentiation logic for finding acceleration from equation 1.
Formula used from calculating acceleration/retardation is
$a=\dfrac{dv}{dt}$
Differentiating velocity with respect to time will give the acceleration at that particular time.
So put value of v in above equation, we get
$\Rightarrow a=\dfrac{d(6-0.02t)}{dt}$
$\Rightarrow a=\dfrac{d(6)}{dt}-\dfrac{d(0.02t)}{dt}$.
On simplifying it using the rules of differentiation ,we get
$\therefore a=-0.02m{{s}^{-2}}$.
So we get the desired value of acceleration.
Since acceleration value is independent of time t so it is uniform acceleration of -0.02ms-2 or uniform retardation of 0.02ms-2.

So, the correct answer is “Option A and D”.

Note: From displacement time equation we can find only velocity equation by differentiating it with respect to time. From the velocity time equation we can find displacement by integrating velocity with respect to time as well as acceleration by differentiating velocity with respect to time. From the acceleration time equation we can find velocity by integrating acceleration with respect to time.