# Question

At 10.00am, Mr Koh’s car passed a certain point, A, travelling at an average speed of 80km/h. At 11.30am, Mr Bob’s car started off from point A at an average speed of 100km/h in pursuit of Mr Koh’s car. a. At what time did Mr Bob’s car overtake Mr Koh’s car?

b. How far was Mr Bob’s car from point A when it overtook Mr Koh’s car?

Speed, S(k) = 80 km/h

Time traveled, T(k) = (t + 1.5) hrs  [Mr Koh had a 1.5 hour head start]

Distance traveled, D(k) = S(k) x T(k) = 80 (t + 1.5) = (80t + 120) km

Speed, S(b) = 100 km/h

Time traveled, T(b) = t hrs

Distance traveled, D(b) = S(b) x T(b) = 100t km

When Mr Bob overtook Mr Koh, they had traveled the same distance, ie, D(k) = D(b)

80t + 120 = 100t

20t = 120

t = 6 hours

(a) 6 hours after 11:30am is 5:30pm

(b) D(b) = 100 x 6 = 600km

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(a)

From 10am to 11.30am, 1.5 h have passed.

Mr Koh would be 80 km/h x 1.5 h = 120 km ahead of Mr Bob.

Mr Bob travelled 100 km/h – 80 km/h = 20 km/h faster than Mr Koh.

Mr Bob would need 120 km ÷ 20 km/h = 6 h to catch up with Mr Koh.

6 h after 11.30am is 5.30pm.

(b)

Mr Bob car would be 100 km/h x 6 h = 600 km away from point A.

Hope it helps 🙂

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