Calculus and Linear Kinematics

## Calculus and Linear Kinematics

The gradient on a distance-time graph gives the velocity, therefore if an equation involves distance and time then it can be differentiated to give velocity.

Similarly, the gradient on a velocity-time graph gives acceleration. Differentiating a velocity-time graph will give acceleration.

## Example 1

A particle is moving between two nodes. The displacement of the particle is given by s = t^3 - 2t^2-12t. What is the velocity of the particle after 5 seconds?

 Displacement s = t^3 - 2t^2 -12t frac(text(d)s)(text(d)t) = velocity frac(text(d)s)(text(d)t) = 3t^2 - 4t -12 v = 3t^2 - 4t -12 At 5 seconds = 3(5)^2 - 4(5) -12 = 43 ms^(-1)

Answer: 43 ms^(-1)

## Example 2

The displacement of a particle is given by s=t^3 - t^2 - t. Determine when the velocity will be zero.

 Displacement s = t^3 - t^2 -t velocity = frac(text(d)s)(text(d)t) = 3t^2 - 2t -1 When is velocity zero? 0  = 3t^2 - 4t -1 Use solution for quadratic x = frac(-b ± sqrt(b^2-4ac))(2a)  x = frac(-(-2) ± sqrt((-2)^2-4(3)(-1)))(2(3))  x = 1

Ignore the negative time value, as it makes no sense in this context.

Answer: After 1 second