Alliance Form 3 P1 Q23 — Speed–Time Graph of a Cheetah

KCSE (Form 3) Form 3 Measurement

Published

The Question

“A speed–time graph describes a cheetah that covered 825 m in 40 seconds. The cheetah starts at 15 m/s, accelerates uniformly to a maximum speed over the first 10 seconds, holds that maximum speed, then decelerates uniformly to rest at 40 seconds (the deceleration lasting the final 20 seconds). Find (a) the speed when timing began, (b) the maximum speed, (c) the acceleration during the first 10 seconds and the deceleration during the last 20 seconds, and (d) the average speed during the first 20 seconds.”

1

Read the starting speed off the axis

The speed when recording started is simply the value where the graph meets the vertical speed axis, that is, the speed at time zero. Reading it directly from the graph gives the answer to part (a) with no calculation needed.

v0=15 m/sv_{0} = 15 \ \text{m/s}
2

Set the total area equal to the distance

On a speed–time graph the distance travelled equals the area under the line. The shape splits into three pieces: a trapezium for the first 10 seconds while the speed rises from 15 to the maximum H, a rectangle while it holds H, and a triangle for the last 20 seconds while it falls to zero. Adding these areas and setting the total equal to 825 gives one equation in H.

12(15+H)(10)+H(10)+12(H)(20)=825\tfrac{1}{2}(15 + H)(10) + H(10) + \tfrac{1}{2}(H)(20) = 825
75+5H+10H+10H=82575 + 5H + 10H + 10H = 825
75+25H=82575 + 25H = 825

The middle rectangle lasts 10 seconds, from t = 10 s to t = 20 s, because the acceleration takes 10 s and the deceleration takes 20 s out of the 40 s total.

3

Solve for the maximum speed H

Subtract 75 from both sides, then divide by 25. This isolates H and gives the top speed the cheetah reached, answering part (b).

25H=75025H = 750
H=30 m/sH = 30 \ \text{m/s}
4

Find the accelerations as gradients

Acceleration is the gradient of a speed–time graph, that is, the change in speed divided by the time taken. For the first 10 seconds the speed climbs from 15 to 30, giving a positive acceleration. For the last 20 seconds it drops from 30 to 0, giving a negative gradient, which is a deceleration.

a1=30−1510=1.5 m/s2a_{1} = \frac{30 - 15}{10} = 1.5 \ \text{m/s}^{2}
a2=0−3020=−1.5 m/s2a_{2} = \frac{0 - 30}{20} = -1.5 \ \text{m/s}^{2}
5

Compute the average speed over the first 20 seconds

Average speed is total distance divided by total time. For the first 20 seconds the distance is the area of the trapezium (first 10 s) plus the rectangle (next 10 s), which is 225 + 300 = 525 m. Dividing by the 20 seconds gives the average speed for part (d).

vˉ=225+30020=52520=26.25 m/s\bar{v} = \frac{225 + 300}{20} = \frac{525}{20} = 26.25 \ \text{m/s}

Final Result

The starting speed is 15 m/s, the maximum speed is 30 m/s, the acceleration in the first 10 seconds is 1.5 m/s squared and the deceleration in the last 20 seconds is 1.5 m/s squared (gradient −1.5), and the average speed over the first 20 seconds is 26.25 m/s.

Why this method works

The method works because the region between a speed–time line and the time axis has a height equal to speed and a width equal to time, so its area carries units of speed multiplied by time, which is distance. Setting that total area equal to the known 825 m turns a geometry picture into a single equation for the unknown height H. In the same way, the steepness of the line measures how fast speed itself is changing, which is exactly the definition of acceleration, so the gradient gives it directly. Average speed then reuses the same area idea over just the part of the journey being asked about.

✓

Check the total: trapezium 225 + rectangle 300 + triangle 300 = 825 m in 40 s, matching the given distance.