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Question: A driver, having a definite reaction time, is capable of stopping over a distance of \(30\,m\) on se...

A driver, having a definite reaction time, is capable of stopping over a distance of 30 m30\,m on seeing a red traffic light when speed of the car is 72 kmhr−172\,kmh{r^{ - 1}} and over a distance of 10 m10\,m when the speed of the car is 36 kmhr−136\,kmh{r^{ - 1}}. Assuming that his reaction time and the acceleration of the car remain the same in all cases, Find the distance over which he can stop the car when he is going at 54 kmhr−154\,kmh{r^{ - 1}}.
(A) 11.75 m11.75\,m
(B) 13.75 m13.75\,m
(C) 15.75 m15.75\,m
(D) 18.75 m18.75\,m

Explanation

Solution

Initially the distance travelled by the car, after the diver has seen the signal and before the driver applies the brake can be determined by distance, velocity and time relation. And then, the acceleration of the distance between 30 m30\,m and 20 m20\,m can be determined. Then, the acceleration of the can for remaining 10 m10\,m distance can be determined. By equating the two acceleration values, the time can be determined, by using this time and the velocity given in the question, the distance can be determined.

Useful formula:
The velocity, distance and time relation is given as,
v=dtv = \dfrac{d}{t}
Where, vv is the velocity, dd is the distance and tt is the time.
The equation of motion is given as,
a=(u2−v2)2sa = \dfrac{{\left( {{u^2} - {v^2}} \right)}}{{2s}}
Where, aa is the acceleration, uu is the initial velocity,vv is the final velocity and ss is the displacement.

Complete step by step solution:
Given that,
The driver is capable of stopping over a distance of 30 m30\,m on seeing a red traffic light when speed of the car is 72 kmhr−172\,kmh{r^{ - 1}},
A distance of 10 m10\,m when the speed of the car is 36 kmhr−136\,kmh{r^{ - 1}}.
The velocity of the given speed is,
72 kmhr−1=20 ms−1 54 kmhr−1=15 ms−1 36 kmhr−1=10 ms−1  72\,kmh{r^{ - 1}} = 20\,m{s^{ - 1}} \\\ 54\,kmh{r^{ - 1}} = 15\,m{s^{ - 1}} \\\ 36\,kmh{r^{ - 1}} = 10\,m{s^{ - 1}} \\\
Now,
The distance travelled by the car at 72 kmhr−172\,kmh{r^{ - 1}} is,
d=v×td = v \times t
On substituting the values, then
d=72×518td = 72 \times \dfrac{5}{{18}}t
Here the unit of speed is converted from kilometre per hour to metre per second, then
The distance travelled is (20×t) m\left( {20 \times t} \right)\,m
Thus, the brakes stopped the car within the distance of (30−20t) m\left( {30 - 20t} \right)\,m, when the speed is at 72 kmhr−172\,kmh{r^{ - 1}} is,
a=(02−202)2×(30−20t)a = \dfrac{{\left( {{0^2} - {{20}^2}} \right)}}{{2 \times \left( {30 - 20t} \right)}}
Here, the initial velocity is zero and the final speed of 72 kmhr−172\,kmh{r^{ - 1}} is substituted in velocity term of 20 ms−120\,m{s^{ - 1}}, then
a=−(202)2(30−20t)a = \dfrac{{ - \left( {{{20}^2}} \right)}}{{2\left( {30 - 20t} \right)}}
By taking 1010 as the common term in denominator, then,
a=−(202)20(3−2t)a = \dfrac{{ - \left( {{{20}^2}} \right)}}{{20\left( {3 - 2t} \right)}}
By cancelling the terms in numerator and denominator, then,
a=−20(3−2t) .............(1)a = \dfrac{{ - 20}}{{\left( {3 - 2t} \right)}}\,.............\left( 1 \right)
Now,
The distance travelled by the car at 36 kmhr−136\,kmh{r^{ - 1}} is,
d=v×td = v \times t
On substituting the values, then
d=36×518td = 36 \times \dfrac{5}{{18}}t
Here the unit of speed is converted from kilometre per hour to metre per second, then
The distance travelled is (10×t) m\left( {10 \times t} \right)\,m
Thus, the brakes stopped the car within the distance of (10−10t) m\left( {10 - 10t} \right)\,m, when the speed is at 36 kmhr−136\,kmh{r^{ - 1}} is,
a=(02−102)2×(10−10t)a = \dfrac{{\left( {{0^2} - {{10}^2}} \right)}}{{2 \times \left( {10 - 10t} \right)}}
Here, the initial velocity is zero and the final speed of 36 kmhr−136\,kmh{r^{ - 1}} is substituted in velocity term of 10 ms−110\,m{s^{ - 1}}, then
a=−(102)2(10−10t)a = \dfrac{{ - \left( {{{10}^2}} \right)}}{{2\left( {10 - 10t} \right)}}
By taking 1010 as the common term in denominator, then,
a=−(102)2×10(1−t)a = \dfrac{{ - \left( {{{10}^2}} \right)}}{{2 \times 10\left( {1 - t} \right)}}
By cancelling the terms in numerator and denominator, then,
a=−102(1−t)a = \dfrac{{ - 10}}{{2\left( {1 - t} \right)}}
On dividing the above equation, then
a=−5(1−t) ...............(2)a = \dfrac{{ - 5}}{{\left( {1 - t} \right)}}\,...............\left( 2 \right)
Now,
The distance travelled by the car at 54 kmhr−154\,kmh{r^{ - 1}} is,
d=v×td = v \times t
On substituting the values, then
d=54×518td = 54 \times \dfrac{5}{{18}}t
Here the unit of speed is converted from kilometre per hour to metre per second, then
The distance travelled is (15×t) m\left( {15 \times t} \right)\,m
Thus, the brakes stopped the car within the distance of (x−15t) m\left( {x - 15t} \right)\,m, when the speed is at 54 kmhr−154\,kmh{r^{ - 1}} is,
Here, xx is the stopping distance, when the car is travelling at 54 kmhr−154\,kmh{r^{ - 1}}
a=(02−152)2×(x−15t)a = \dfrac{{\left( {{0^2} - {{15}^2}} \right)}}{{2 \times \left( {x - 15t} \right)}}
Here, the initial velocity is zero and the final speed of 54 kmhr−154\,kmh{r^{ - 1}} is substituted in velocity term of 15 ms−115\,m{s^{ - 1}}, then
a=−(152)2(x−15t) .....................(3)a = \dfrac{{ - \left( {{{15}^2}} \right)}}{{2\left( {x - 15t} \right)}}\,.....................\left( 3 \right)
It is given that the acceleration is same for all cases, then by equating the equation (1) and equation (2), then
−20(3−2t)=−5(1−t)\dfrac{{ - 20}}{{\left( {3 - 2t} \right)}} = \dfrac{{ - 5}}{{\left( {1 - t} \right)}}
The negative sing gets cancelled on both sides, then
20(3−2t)=5(1−t)\dfrac{{20}}{{\left( {3 - 2t} \right)}} = \dfrac{5}{{\left( {1 - t} \right)}}
By rearranging the terms, then
20(1−t)=5(3−2t)20\left( {1 - t} \right) = 5\left( {3 - 2t} \right)
By multiplying the above equation, then
20−20t=15−10t20 - 20t = 15 - 10t
By keeping the term tt in one side and the other terms in other side, then
−10t+20t=20−15- 10t + 20t = 20 - 15
On further calculation,
10t=510t = 5
By keeping the term tt in one side and the other terms in other side, then
t=510t = \dfrac{5}{{10}}
On dividing the above equation, then
t=0.5 st = 0.5\,s
By equating the equation (2) and equation (3), then
−5(1−t)=−(152)2(x−15t)\dfrac{{ - 5}}{{\left( {1 - t} \right)}} = \dfrac{{ - \left( {{{15}^2}} \right)}}{{2\left( {x - 15t} \right)}}
The negative sing gets cancelled on both sides, then
5(1−t)=(152)2(x−15t)\dfrac{5}{{\left( {1 - t} \right)}} = \dfrac{{\left( {{{15}^2}} \right)}}{{2\left( {x - 15t} \right)}}
By rearranging the terms, then
5×2(x−15t)=152×(1−t)5 \times 2\left( {x - 15t} \right) = {15^2} \times \left( {1 - t} \right)
By substituting the tt value in the above equation, then
5×2(x−(15×0.5))=152×(1−0.5)5 \times 2\left( {x - \left( {15 \times 0.5} \right)} \right) = {15^2} \times \left( {1 - 0.5} \right)
On further calculation, then
10(x−7.5)=225×0.510\left( {x - 7.5} \right) = 225 \times 0.5
By keeping the term xx in one side and the other terms in other side, then
(x−7.5)=225×0.510\left( {x - 7.5} \right) = \dfrac{{225 \times 0.5}}{{10}}
On further calculation in RHS, then
(x−7.5)=11.25\left( {x - 7.5} \right) = 11.25
By keeping the term xx in one side and the other terms in other side, then
x=11.25+7.5x = 11.25 + 7.5
By adding the terms, then
x=18.75 mx = 18.75\,m
Thus, the above equation shows the Find the distance over which he can stop the car when he is going at 54 kmhr−154\,kmh{r^{ - 1}}.

Hence, the option (D) is correct.

Note: In this question, it is given that the acceleration is the same for all the three conditions, so the acceleration equation can be determined and equated with one another. So, the time taken to stop and the distance covered by the car at 54 kmhr−154\,kmh{r^{ - 1}} can be determined.