Question

The graph below shows how the velocity of a motorbike
varies with time during the final 10s of a race.
(i)Describe the motion shown by the graph.
(ii)Calculate the acceleration in the first phase of the motion.
(iii)Show that during the final 10 s the motorbike travels a distance of approximately 800 m

Answer 1

Explanation:

I see an object that is has a Positve acceleration for 4 seconds then a constant acceleration for the final 6 seconds.

ii. Acceleration is the slope of a velocity versus time graph.

Since the first phase is a line, it has a uniform slope.

So we need to chose two points to find the acceleration

Let's use (2,60) and (4,90).

$a = \frac{90 - 60}{4 - 2} = \frac{30}{2} = 15$

So the acceleration is 15.

This means, for any velocity versus time graph, the velocity is

$\frac{v _{f} - v _{i}}{t} = a$

This is velocity final- velocity initial divided by time equals Acceleration

iii. The average velocity is the the sum of the final velocity and initial velocity divided by 2.

So the average velocity is

$\frac{30 + 90}{2} = 60$

Then to find distance

$v _{avg}t = d$

So during the first interval, the motorbike travels

$60 \times (4) = 240$

In the next part, the average velocity stays at 90 and this occurs for 6 seconds

$90 \times 6 = 540$

Next, we add the velocities

$240 + 540 = 780$

So the distance travleed is 780 meters which is close to 800