# Kinematics V- Calculus Based (Acceleration)

### Introduction

Acceleration is a very important quantity while studying about the kinematics of the particle. The rate of change of velocity with respect to time is acceleration. It is denoted by a-bar (alphabet “a” with a bar on top, as it is the vector quantity).

a bar =_{ }( V_{f} – V_{i} )/ t

The above was average acceleration. If you want to find the acceleration at an instant i.e. instantaneous acceleration, then we must use calculus. To find instantaneous acceleration, we calculate the first order derivative of velocity with respect to time, a_{w}= dv/dt. Acceleration can also be obtained by the second order derivative of position with respect to time. a_{w}= (d^{2}x) / dt^{2}.

### When is acceleration zero (on graphs)

The slope of a velocity/time graph gives us the acceleration of the body. The slope of the graph at a point will give instantaneous acceleration. The acceleration of the body can be zero in 3 cases. that are at the point of maximum velocity at the point of the minimal velocity and when the velocity is constant. At maxima and minima, you get a parallel line to x-axis which marks an angle 0^{0} with x axis and slope tan theta => tan 0= 0 therefore, the slope is zero (look at image). And when the body move with uniform/constant velocity then the derivative of it becomes zero as the derivatives constant is zero.

### Practice

**Example problem: **The position of an object as a function of time is given as x(t) = 3t^3+8t. Find velocity and acceleration at t=3.

**Solution: **To find the function of velocity, we differentiate the function x(t)

V(t)= dx/dt = d( 3t^3+8t ) / dt =** 9t^2+8**

To find v at t=3, substitute 3 in place of t in v(t).

the answer is- 9 (3)^2 + 8 = 81+8 = **90 m/s**

To find function of acceleration, differentiate v(t) with respect to time.

the answer is 18t+0 = **18t **

At t=3, acceleration is **54 m/s^2**