Further Variable Acceleration

Example 1:

A particle is moving in a straight line with acceleration, at time seconds, given by

ms

The velocity of the particle at time is ms.

a) Find an expression for the velocity at time seconds.

Step 1: To find the velocity, we need to integrate the expression for the acceleration.

Step 2: Use integration by recognition to make a guess as to what may differentiate to get .

We will guess . The angle is , which differentiates to , differentiates to , and then the angle inside stays the same.

Angle stays the same



Differentiated angle

Our guess differentiates to give , but we wanted . So, we multiply both sides the constant we want over the constant we have, which will be .

Step 3: Find using the initial conditions when given in the question.

ms

b) Find the maximum and minimum speed.

Single Step: The maximum value of occurs when and the minimum occurs when .

ms

ms

c) Find the distance travelled in the first seconds.

Step 1: To find the distance, we need to integrate the velocity between and seconds.

First, we need to ensure that the velocity is positive within that domain. From part b), the minimum value of is , which is greater than , so is always positive.

Take out of the integral to leave us with an easier expression to integrate

Step 2: Use integration by recognition to make a guess as to what may differentiate to get .

We will guess . The angle is , which differentiates to , differentiates to , and then the angle inside stays the same. We know that integrates to .

Angle stays the same



Differentiated angle

Our guess differentiates to , but we wanted . So, we multiply both sides by the constant we want over the constant we have, which will be . So, we have:

Step 3: Substitute the limits and to find the distance travelled.

m

Example 2:

A particle of mass kg is moving on the positive axis. At time seconds, the displacement, m, of the particle from the origin is given by

a) Find the velocity of the particle when .

Single Step: To find the velocity, we need to differentiate the expression for displacement.

Substituting , we get:

ms

The particle is acted on by a single force of variable magnitude N, which acts in the direction of the positive axis.

b) Find the magnitude of when .

Step 1: We’ll use Newton’s second law here, namely . To find the acceleration, we need to differentiate the equation for velocity.

Substituting , we get:

ms

Step 2: Substitute the values of the mass and acceleration into the formula to obtain the force .

N

Since we want the magnitude, we are finding .

N

N