Variable Acceleration

1. A racing car accelerates from rest along the track. Its speed, , is modelled for the first four seconds of its motion by

(a) Find an expression for the distance travelled by the car in the first seconds.

 Hint

 Solutions

The speed is given by

The distance travelled in the first seconds is

Thus

with in metres.

(b) Calculate the distance travelled from to .

 Hint

 Solutions

Recall

Hence

and

The distance travelled from to is

Show that the acceleration, , of the car at time is given by , where is a constant to be determined.

 Hint

 Solutions

Differentiate v with respect to t:

Factor out 3 and factor the quadratic:

Hence k=3.

2. A particle moves on the -axis. Its displacement, , from the origin O is given by , where is the time in seconds.

How far is the particle from O when it is instantaneously at rest?

 Hint

Get the velocity by differentiating the displacement first. Then set the velocity to zero to find the time when the particle is at rest. Finally, substitute this time back into the displacement equation to find how far the particle is from O.

 Solutions

Differentiate with respect to to find the velocity :

Set to find when the particle is instantaneously at rest:

Substitute into the equation for to find the displacement from O:

The particle is from O when it is instantaneously at rest.

3. At a particle travels through point P with a velocity of . The acceleration of the particle is given by .
   (a) Find the times when the velocity, , is .

 Hint

For variable acceleration, integrate the acceleration function to find the velocity function. Use the initial velocity condition to find the constant of integration. Then set the velocity function equal to -18 and solve for t.

 Solutions

Integrate the acceleration function to find the velocity function:

Use the initial condition when to find :

Thus, the velocity function is:

Set and solve for :

Rearranging gives:

Multiply through by 10 to eliminate decimals:

Using the quadratic formula:

Thus, the two solutions for are:

(b) Find the distance covered by the particle when .

 Hint

To find the distance covered when , first determine the time intervals when this condition holds. Then, integrate the velocity function over these intervals to find the total distance traveled.
Basically, the distance covered is the area under the velocity-time graph between the times when .

 Solutions

From part (a), we found that at and .
To find the distance covered when , we need to integrate the velocity function from to .
The velocity function is:

Integrate the velocity function to find the displacement:

We can ignore the constant of integration since we are calculating a definite integral. C will get canceled out.

Now, calculate the displacement from to :

Calculating each term:
For :

For :

Now, find the distance covered, the area is under the x axis so it will be negative, we make it positive for distance:

The distance covered by the particle when is .

4. An insect moves in a straight line. The time, , is in seconds and distance travelled is in metres.

The velocity, , of the insect is given by

You are also given that when .
(a) Show that .

 Hint

C is the constant velocity between and . It is also the velocity at from the first equation. So, substitute into the first equation to find .

 Solutions

to find , we substitute into the first equation for velocity:

Thus,

(b) Calculate the values of and and briefly describe the motion of the insect in the interval .

 Hint

Need to setup two equations using the information given. One equation comes from the fact that the velocity at must equal , and the other comes from the fact that the velocity at is given as 4. Solve these two equations simultaneously to find and .

 Solutions

From part (a), we have .
At , the velocity from the third equation must equal :

At , the velocity is given as 4:

Subtract equation (1) from equation (2):

Substituting into equation (1):

Thus, and .
The motion of the insect in the interval is described by the equation:

This indicates that the insect is decelerating linearly over this time period. The decelerating is constant.

Sketch the - curve for the motion of the insect in the interval .

 Hint

quadratic curve in the interval .

 Solutions

(d) Calculate the distance travelled by the insect in the interval .

 Hint

The distance travelled is the area under the velocity-time graph from to .
There are two parts to this area: from to (which is a curve) and from to (which is a rectangle).
First part can be found by integrating the velocity function from to . But since between and the velocity is negative, we need to split the integral into two parts: from to and from to .

 Solutions

Since between and the velocity is negative, we need to split the integral into two parts: from to and from to .
First, calculate the distance from to :
integrate the velocity function:

The distance is the absolute value:

Next, calculate the distance from to :

Now, calculate the distance from to :
The velocity is constant at :

The total distance travelled from to is:

5. The motion of a particle is represented on the velocity time graph shown below. The motion is given by the equations:

The particle returns to its starting position at time .
Find the velocity of the particle at time .

 Hint

The particle returns to its starting position when the total displacement is zero. So we need to find the expressions for displacement by integrating the velocity function. Then, set the total displacement to zero and solve for .

Then substitute back into the velocity equation to find the velocity at time .

There are two parts to the displacement: from to and from to . Will calculate each part separately.

 Solutions

First compute the displacement from 0 to 4:

Displacement is negative, showing in the graph that the area are under x axis.

Continuity of velocity at t=4 gives

so

and for we have

Let be the time the particle returns to the start. For the particle to return to original position, the integration (area) between 4 and T need to be:

Evaluating the integral gives

hence

The relevant root (with ) is

Finally the velocity at T is

Numerically v(T)≈7.86 ms⁻¹.