Question

Two cars A and B with velocity 280m/s and 80m/s respectively, Car B

leaves point X 6:00am and car A also leaves the same point 6:40am
calculate the distance between the two cars after [I] 1hr [ii] where do you
think car A will overtake car B ​

Answer 1

Car B travels a distance x after time t (measured in seconds) of

x = (80 m/s) t

so that at 6:40 AM (which is 40 min = 2400 s after the starting time of 6:00 AM), it will have moved

(80 m/s) * (2400 s) = 192,000 m

If we take t = 0 to refer to 6:40 AM, the cars' respective positions x at time t are given by

A: x = (280 m/s) t

B: x = 192,000 m + (80 m/s) t

(i) After 1 h = 3600 s, their positions are

A: (280 m/s) * (3600 s) = 1,008,000 m

B: 192,000 m + (80 m/s) * (3600 s) = 480,000 m

away from the point X, so the distance between the two cars is

1,008,000 m - 480,000 m = 528,000 m = 582 km

(ii) Car A moves faster than car B, so A overtakes B as soon as their positions are equal, which happens for

(280 m/s) t = 192,000 m + (80 m/s) t

(200 m/s) t = 192,000 m

t = (192,000 m) / (200 m/s)

t = 960 s = 16 min

So after 16 min, both cars have a position of

(280 m/s) * (960 s) = 268,800 m = 268.8 km

beyond point X.