A drunken motorist who is moving at a constant velocity passes a stationary police patrol car. The patrol car immediately gives chase at a constant acceleration and catches up with the motorist after a distance of 15 km in 450 s. Below is the graph of velocity against time for both vehicles. Calculate the speed V₁ and V₂ when the patrol car catches up with the motorist. * 1 point Captionless Image
Answers
Answer:
1/6hr or 0.1667 minutes
Explanation:
Start counting time t from when the drunk motorist passes. Let his distance travelled be X1 and the distance travelled by the police car be X2.
X1 = V*t
X2 = (a/2) t^2
X1 = X2 when both equal 15 km. That is when the police car catches up.
t = 15 km/(90 km/h) = 1/6 h = 0.1667 minutes
You could also solve for the police car's acceleration a, but they don't ask for that.
Given : A drunken motorist who is moving at a constant velocity passes a stationary police patrol car.
The patrol car immediately gives chase at a constant acceleration and catches up with the motorist after a distance of 15 km in 450 s.
To Find : the speed V₁ and V₂ when the patrol car catches up with the motorist
Solution:
Distance covered by motorist = V₁ * t ( area of rectangle)
t = 450 sec
Distance = 15 km = 15000 m
=> 15000 = V₁ * 450
=> V₁ = 15000/450
=> V₁ = 100/3
=> V₁ = 33.33 m/s
Distance covered by police patrol car. ( area of triangle )
= (1/2) * V₂ * t
= (1/2) * V₂ * 450
= V₂ * 225
=> 15000 = V₂ * 225
=> V₂ = 200/3
=> V₂ = 66.67 m/s
Hence correct answer is
V₁ = 33.33 m/s and V₂ = 66.67 m/s
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