Question

Jim is driving his car at 32 m/s (72 mi/h) along a highway where the speed limit is 25 m/s (55 mi/h). A highway patrol car observes him pass and quickly reaches a speed of 35 m/s. At that point, Jim is 350 m ahead of the patrol car. How far does the police car travel before catching Jim?

Answer 1

Answer:

The police car travel before catching Jim is 4.08 km.

Explanation:

Given that,

Distance = 350 m

Speed of car = 32 m/s

Speed of petrol car = 35 m/s

Patrol car covered the distance is 350+x with the speed of 35 m/s.

Car covered the distance is x with the speed of 32 m/s.

We need to calculate the time for patrol car

$t=\dfrac{35}{350+x}$

We need to calculate the time for car

$t=\dfrac{32}{x}$

The time of both car is same.

$\dfrac{35}{350+x}=\dfrac{32}{x}$

$x = \dfrac{350\times32}{3}$

$x=3733.33\ m$

We need to calculate the police car travel before catching Jim

$x'=x+350$

$x'=3733.33+350$

$x'=4083.33\ m$

$x'=4.08\ km$

Hence,  The police car travel before catching Jim is 4.08 km.

Answer 2

Answer:

Given:

v1=32m/s,

v2=35m/s,

d1=250m,

d2=?

If both cars are travelling in the same direction, one at 32ms-1 and the other at 35 ms-1 then their relative velocity is Vr 3ms-1(by vector addition).

The time to catching Jim is t=d1/Vr= 250/3 = 83.3s

The distance traveled by the patrol car travel before catching Jim is d2 = v2t = v2 d1/Vr =35∙250/3= 2916.7m

Answer: d2= 2916.7m

Explanation: