Two students, P and Q, run along a straight track. Fig. 10.1 shows the distance-time graphs for both students. Student P starts at time t = 0 and runs at a ...
Question 1 Report
Two students, P and Q, run along a straight track. Fig. 10.1 shows the distance-time graphs for both students. Student P starts at time t = 0 and runs at a constant speed. Student Q starts 5.0 s later and also runs at a constant speed, but faster than P. Both students run in the same direction from the same starting line. The track is 150 m long. A teacher uses two stopwatches to time each student independently and records the results on a whiteboard for the class to analyse.
(a) Calculate the speed of student P. [2]
(b) Calculate the speed of student Q. [2]
(c) Determine the time at which Q overtakes P. [2]
(d) State the distance from the start at which Q overtakes P. [1]
Answer Details
(a) Speed of student P
From the graph, student P starts at t = 0 and reaches 150 m at t = 30 s. Since P travels at constant speed, the distance-time graph is a straight line.
\(\text{speed of P} = \frac{\text{distance}}{\text{time}} = \frac{150}{30}\) [1]
= 5.0 m/s [1]
(b) Speed of student Q
Student Q starts at t = 5.0 s and reaches 150 m at t = 25 s. The time Q is running = 25 - 5 = 20 s.
\(\text{speed of Q} = \frac{150}{20}\) [1]
= 7.5 m/s [1]
(c) Time at which Q overtakes P
At the moment of overtaking, both students have covered the same distance from the start line.
- Distance covered by P at time t: \(d_P = 5.0t\)
- Distance covered by Q at time t (Q started at t = 5.0 s): \(d_Q = 7.5(t - 5)\)
Setting \(d_P = d_Q\):
\(5.0t = 7.5(t - 5)\)
\(5.0t = 7.5t - 37.5\)
\(37.5 = 2.5t\) [1]
\(t = 15\) s [1]
(d) Distance from the start at which Q overtakes P
\(d = 5.0 \times 15 = \) 75 m [1]
This can be confirmed from Q's equation: \(7.5 \times (15 - 5) = 7.5 \times 10 = 75\) m. Both answers agree, confirming the overtake point.
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