The diagram shows a velocity-time graph of a motorbike for 25 seconds. a) After how many seconds was the acceleration zero? 4.5 seconds
metres
b) What was the distance travelled in the last 15 seconds?
25
20
15
Velocity (m/s)
10
5-
0
0
5
10 Time (s)15
20
2
25

 The acceleration is 0 after 4.5 seconds.

*The distance traveled in the last 15 seconds on the graph is about 225 meters.*

If the question means "distance traveled during the last 15 seconds before coming to a complete stop" The distance is 30 meters.

The attachment shows what I believe to be the missing right side of the graph showing the velocity at 0.

Step-by-step explanation: a.) at 4.5 seconds, the graph is momentarily flat; that is, the derivative, (the change in slope) is 0. There is no change in velocity at that point. Before that, the velocity is increasing, so there is positive acceleration. After that, the velocity is decreasing. So the acceleration is negative.

*The answer choices seem incorrect to me. for part b. Is there a typo?*

b.) From 10 seconds to 25 seconds, the "last 15 seconds" on the graph, the velocity is decreasing at a steady rate; the average velocity is 15m/sec

At  15m/sec × 15 sec, the distance traveled is 225 meters.

If we are to extrapolate to the point there the velocity is 0, assuming the steady rate of deceleration, the motorbike will reach that stopping point at 63.75 seconds. At 15 seconds before that, at 48.5 sec, it is traveling at 4m/sec.

The average velocity during the final 15 seconds is 2m/sec.

2m/sec × 15 sec = 30 meters.

I don't see that choice either, so I wonder if the question is missing something.

If someone else has a better explanation, I hope they they will answer and explain what I missed.

y - y1 = m(x- x1)

y - (-5) = 2/3(x -0)

y + 5 = 2/3x

y = 2/3x - 5