Two racing cars, P and Q, start a drag race from rest on a straight track. Fig. 1.1 shows the speed-time graphs for both cars during the first 10 s of the r...

(a) Acceleration of car P

Car P accelerates uniformly from rest (\(u = 0\)) to 40 m/s in 8.0 s:

\(a = \frac{v - u}{t} = \frac{40}{8.0} = 5.0\) m/s² [1]

(b) Acceleration of car Q

Car Q accelerates uniformly from rest to 50 m/s in 10 s:

\(a = \frac{50}{10} = 5.0\) m/s² [1]

Both cars have the same acceleration, but car Q reaches a higher final speed because it accelerates for a longer time.

(c) Distance travelled by car P in the first 10 s

Car P's journey in 10 s has two phases:

1. Phase 1 (0 to 8.0 s): Uniform acceleration from rest to 40 m/s. The distance is the area of the triangle on the speed-time graph:
   \(d_1 = \frac{1}{2} \times 8.0 \times 40 = 160\) m
2. Phase 2 (8.0 to 10 s): Constant speed of 40 m/s for 2.0 s:
   \(d_2 = 40 \times 2.0 = 80\) m

Total: \(d = 160 + 80\) [1] \(= 240\) m [1]

(d) Distance travelled by car Q in the first 10 s

Car Q accelerates uniformly from rest to 50 m/s over the entire 10 s. The distance is the area of the triangle under its speed-time graph:

\(d = \frac{1}{2} \times 10 \times 50\) [1] \(= 250\) m [1]

(e) Which car is ahead at t = 10 s and by how much

Car Q has travelled 250 m and car P has travelled 240 m. Since both started from the same position:

Car Q is ahead by \(250 - 240 = 10\) m [1]

Despite having the same acceleration (5.0 m/s²), car Q is ahead because it accelerated for longer (10 s vs 8.0 s) and reached a higher top speed. On a speed-time graph, the distance is represented by the area under the curve, and Q's triangle has a larger area than P's triangle-plus-rectangle combination over this interval.