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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=t3-2t2-12t. What is the velocity of the particle after 5 seconds?

Displacement s =t3-2t2-12t
dsdt = velocity dsdt =3t2-4t-12
v =3t2-4t-12
At 5 seconds =3(5)2-4(5)-12
=43ms-1

Answer: 43ms-1

Example 2

The displacement of a particle is given by s=t3-t2-t. Determine when the velocity will be zero.

Displacement s =t3-t2-t
velocity = dsdt =3t2-2t-1
When is velocity zero? 0 =3t2-4t-1
Use solution for quadratic x =-b±b2-4ac2a
x =-(-2)±(-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