The shortest possible distance in which she catches up with the man is 129.7 m.
The given parameters:
speed of the man, v₁ = 40 km/hr
speed of the woman, v₂ = 70 km/hr
time when the woman started moving, t₁ = 5 seconds
To find:
- the shortest distance when she catches up with the man
Convert the given speed in km/h to m/s
- 3.6 km/h = 1 m/s
- 40 km/h = 40/3.6 = 11.11 m/s
- 70 km/h = 70/3.6 = 19.44 m/s
Calculate the distance moved by the man before she started following him.
Distance moved by the man = speed x time
= 11.11 m/s x 5 s
= 55.55 m
Let the time the woman catches up with the man = t
Apply the following simple principle to calculate the time.
The man is moving forward and the woman is moving forward to close the gap between them, thus both speeds are working against each other.
[tex](v_2-v_1)t = 55.55 \ m[/tex]
- the difference in the speed is because they are working against each other.
[tex](19.44 - 11.11)t = 55.55\\\\8.33t = 55.55\\\\t = \frac{55.55}{8.33} \\\\t = 6.67 \ s[/tex]
The time when the woman catches up with the man = 6.67 s
The shortest distance at which the woman catches up with the man is calculated as;
Shortest distance = woman's speed x time when she catches up
= 19.44 m/s x 6.67 s
= 129.7 m
Thus, the shortest possible distance in which she catches up with the man is 129.7 m.
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