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Key idea:
On a speed–time graph, distance travelled is the area under the graph.
Sam’s distance:
distance = 10t
Joe waits for 4 s, then accelerates from 4 s to 10 s.
Joe’s acceleration:
15 ÷ 6 = 2.5 m/s²
Meaning of catch-up
Joe catches Sam when they are at the same position on the straight road.
That means:
Sam’s distance = Joe’s distance
From 0 to 10 s:
Sam travels 10 × 10 = 100 m.
Joe’s distance from 4 s to 10 s is the triangle area:
½ × 6 × 15 = 45 m.
At 10 s, Sam is ahead by:
100 − 45 = 55 m.
After 10 s:
Sam moves at 10 m/s.
Joe moves at 15 m/s.
Joe gains on Sam at:
15 − 10 = 5 m/s.
Time needed after 10 s:
55 ÷ 5 = 11 s.
Therefore catch-up time from the start:
10 + 11 = 21 s.
Since Joe starts cycling at 4 s, the time after Joe starts is:
21 − 4 = 17 s.
So both answers can make sense depending on what the question means:
21 s from the beginning of the whole situation.
17 s after Joe starts cycling.