Answer :
Answer:
It will take 40 seconds to catch the speeding car
Explanation:
Initial speed of the patrolman = 55 mph
after one minute speed of patrol man = 115 mph
now acceleration of patrolman is given by
[tex]a = \frac{v_f - v_i}{t}[/tex]
[tex]a = \frac{115 - 55}{1/60} = 3600 m/h^2[/tex]
now at this acceleration the distance covered by patrolman in "t" time is given as
[tex]d = v_i t + \frac{1}{2}at^2[/tex]
[tex]d = 55t + 1800t^2[/tex]
now we know the speed of the speeding car is given as
[tex]v' = (55+20) mph[/tex]
now in the same time distance covered by it
[tex]d = 75 t[/tex]
now since the distance covered is same
[tex]75 t = 55t + 1800 t^2[/tex]
[tex]t = \frac{20}{1800} = \frac{1}{90} h[/tex]
[tex]t = 40 seconds[/tex]
(a) After [tex]\boxed{30\text{ s}}[/tex] of speeding up a highway patrolman catches up the speeding car.
(b) After [tex]\boxed{\dfrac{{20}}{{\left( {v - 75} \right)}}{\text{min}}}[/tex] of speeding up a highway patrolman catches up the speeding car.
Further explanation:
Part (a)
If a patrolman and car is going in the same direction, relative velocity of both the cars will be the difference between the velocities of the car.
The patrolman is travelling with the speed limit and is passed by a car going [tex]20\text{ mph}[/tex] faster than the speed limit. The speed limit is [tex]55\text{ mph}[/tex]. So, patrolman is passed by a car with a relative speed of [tex]20\text{ mph}[/tex].
The speed of the car is:
[tex](55+20)=75\text{ mph}[/tex]
After [tex]1\text{ min}[/tex] patrolman speeds up to [tex]115\text{ mph}[/tex].
Therefore, the required required to catch up the fast moving car is equal to the difference between speed of patrolman's car and speed of car.
Required speed is:
[tex]115-75\\=40\text{ mph}[/tex]
Given:
The speed limit for a car is [tex]55\text{ mph}[/tex] .
The speed of the patrolman's car is [tex]115\text{ mph}[/tex].
Concept:
The distance traveled by patrolman with relative to the car is given by the following relation .
[tex]d=v \cdot t[/tex]
Rearrange the above expression for [tex]t[/tex].
[tex]\boxed{t=\dfrac{d}{v}}[/tex] …… (1)
Here, [tex]t[/tex] is the time required to catch up with the car, [tex]d[/tex] is the catch up distance and [tex]v[/tex] is the catch up speed.
The catch up distance is the distance traveled by the car in [tex]1\text{ min}[/tex] with excess speed.
The distance traveled by the patrolman's car in order to catch up the car is:
[tex]\begin{aligned}d&=\left( {20\,{\text{mph}}} \right)\left( {1\,{\text{min}}} \right) \\&=0.333\,{\text{miles}} \\ \end{aligned}[/tex]
Substitute [tex]0.333\text{ miles}[/tex] for [tex]d[/tex] and [tex]40\text{ mph}[/tex] for [tex]v[/tex] in equation (1).
[tex]\begin{aligned}t&=\frac{{0.33\,{\text{miles}}}}{{40\,{\text{miles/h}}}} \\&=\frac{{0.33}}{{40}}{\text{h}}\left( {\frac{{3600\,{\text{s}}}}{{1\,{\text{h}}}}} \right) \\&=30\,{\text{s}} \\ \end{aligned}[/tex]
Thus, after [tex]\boxed{30\text{ s}}[/tex] of speeding up a highway patrolman catches up the speeding car.
Part (b)
Concept:
If patrolman speeds up with the speed of [tex]v[/tex] then catch-up time can be calculated by using equation (1).
In this case, the catch-up speed of the patrolman's car will be [tex](v-75)\text{ mph}[/tex].
Substitute [tex]0.333\text{ miles}[/tex] for [tex]d[/tex] and [tex](v-75)\text{ mph}[/tex] for [tex]v[/tex] in equation (1).
[tex]\begin{aligned}t&=\frac{{0.33\,{\text{miles}}}}{{\left( {v - 75} \right)\,{\text{miles/h}}}} \\&=\frac{{60}}{{3\left( {v - 75} \right)}}{\text{min}} \\&=\frac{20}{(v-75)}\text{ min} \\ \end{aligned}[/tex]
Thus, after [tex]\boxed{\dfrac{{20}}{{\left( {v - 75} \right)}}{\text{min}}}[/tex] of speeding up a highway patrolman catches up the speeding car.
Learn more:
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Answer Details:
Grade: College
Subject: Physics
Chapter: Kinematics
Keywords:
Highway, patrolman, speed limit, 20 mph, faster, one minute, 15 mph, speeding up, catches up, 55 mph, 30 sec, 0.5 minute, 20/(V-70) minutes.
