A bus starts from rest, accelerates uniformly, travels at constant velocity, then decelerates uniformly to rest. Describe the shape of its velocity-time graph.

1. Since the bus starts from rest, the graph begins at the origin.

2. Uniform acceleration is shown by a straight line sloping upwards.

3. Constant velocity is shown by a horizontal line.

4. Uniform deceleration to rest is shown by a straight line sloping down to the time axis.

A tram moves from rest to 18 m s^-1 uniformly in 30 s, then continues at 18 m s^-1 for another 20 s. Find its acceleration in each part.

1. “From rest” means the starting velocity is 0 m s^-1.

2. For the first part, use change in velocity divided by change in time:

   $$a=\frac{18-0}{30}=0.6$$

3. The acceleration for the first 30 s is 0.6 m s^-2.

4. During the constant-velocity part, the graph is horizontal, so the acceleration is 0 m s^-2.

A train starts from rest, reaches 22 m s^-1 in 25 s, travels at this velocity for 40 s, then slows uniformly to rest in 15 s. Find the distance travelled.

1. Split the graph into three areas: a triangle, a rectangle, and another triangle.

2. The first triangle has base 25 and height 22:

   $$A_1=\frac{1}{2}\times 25\times 22=275$$

3. The rectangle has base 40 and height 22:

   $$A_2=40\times 22=880$$

4. The final triangle has base 15 and height 22:

   $$A_3=\frac{1}{2}\times 15\times 22=165$$

5. Add the three areas:

   $$A_1+A_2+A_3=275+880+165=1320$$

6. The train travels 1320 m.

A delivery car starts from rest, accelerates uniformly for 8 s to speed $V$ m s^-1, drives at this speed for 12 s, then decelerates uniformly to rest in 5 s. Describe the labelled sketch.

1. Draw time on the horizontal axis and velocity on the vertical axis.

2. Start at the origin because the car starts from rest.

3. Draw a straight rising line from $(0,0)$ to $(8,V)$.

4. The constant-speed part lasts 12 s, so draw a horizontal line from time 8 to time 20.

5. Draw a straight falling line from $(20,V)$ to $(25,0)$.

6. Label the vertical height $V$ and the times 8, 20 and 25 on the time axis.

A car accelerates from rest to $V$ m s^-1 in 8 s. It maintains this speed for 36 s, then brakes to rest. The magnitude of the deceleration is twice the magnitude of the acceleration. The total distance is 1008 m. Find the total time and the value of $V$.

1. The acceleration magnitude while speeding up is:

   $$a=\frac{V}{8}$$

2. The deceleration magnitude is twice as large:

   $$2a=2\times \frac{V}{8}=\frac{V}{4}$$

3. Find the braking time using change in velocity divided by deceleration magnitude:

   $$t=\frac{V}{\frac{V}{4}}=4$$

4. The total time is:

   $$8+36+4=48$$

5. Now use area under the graph for the distance:

   $$\frac{1}{2}\times 8\times V+36V+\frac{1}{2}\times 4\times V=1008$$

6. Simplify the left-hand side:

   $$4V+36V+2V=1008\Rightarrow 42V=1008$$

7. Solve for $V$:

   $$V=24$$

8. The maximum speed is 24 m s^-1, and the total time is 48 s.

A train starts from rest, accelerates at 0.5 m s^-2 for 40 s, then travels at constant speed before decelerating at 0.4 m s^-2 to rest. The stations are 3.1 km apart. Find the total time.

1. Convert the distance: 3.1 km is 3100 m.

2. Find the top speed after the acceleration phase:

   $$v=0+0.5\times 40=20$$

3. The acceleration section is a triangle with base 40 and height 20:

   $$A_1=\frac{1}{2}\times 40\times 20=400$$

4. Find the deceleration time:

   $$t=\frac{20}{0.4}=50$$

5. The deceleration section is another triangle:

   $$A_3=\frac{1}{2}\times 50\times 20=500$$

6. Find the remaining distance for the constant-speed section:

   $$3100-400-500=2200$$

7. Divide by the constant speed to get the cruising time:

   $$t_c=\frac{2200}{20}=110$$

8. Add all the times:

   $$40+110+50=200$$

9. The total time is 200 s.