Question

You are driving toward a traffic signal when it turns yellow. Your speed is the legal speed limit of ${v_0} = 55\,km/hr$: your best deceleration rate has the magnitude $a = 5.18\,m{s^{ - 2}}$. Your best reaction time to begin braking is $T = 0.75\,s$. To avoid having the front of your car enter the intersection after the light turns red, should you brake to a stop or continue to move at $55\,km/hr$ if the distance to the intersection and the duration of the yellow light are (a) $40\,m\,and\,2.8\,s$, and (b) $32\,m\,and\,1.8\,s$? Give an answer of the brake, continue, either (if either strategy works), or neither (if neither strategy works nor the yellow duration is inappropriate).

Answer 1

Hint: The difference between the final and initial velocity with respect to time is called acceleration. And it is denoted by ‘a’ and SI unit is$m{s^{ - 2}}$. Acceleration is a vector quantity, it means it has direction as well as magnitude. There are two types of acceleration: uniform acceleration and non-uniform accelerations.

Complete step by step answer:
(a) Given, $a = 5.18\,m{s^{ - 2}}$
We take direction of motion as $+x$. so
$a = - 5.18\,m{s^{ - 2}}$
$\Rightarrow T = 0.75s$
$\Rightarrow {v_0} = 55\,km/hr$
We use SI unit
${v_0} = 15.28\,m/s$
The velocity is constant during the reaction time $T$,so the distance travelled during it is,
$\therefore {d_r} = {v_0}T$
Put the value
${d_r} = 15.28 \times 0.75$
$ \Rightarrow {d_r} = 11.46\,m$

The distance travelled during braking,
$\therefore {v^2} = {u^2} + 2as$
Put the value
$s = \dfrac{{15.28}}{{2( - 5.18)}}$
$ \Rightarrow s = 22.53\,m$
Thus, the total distance is ${d_r} + s$, $34\,m < 40\,m$.
It means drivers are able to stop the vehicle in time. And if the driver were to continue at ${V_0}$, the car would enter the intersection in,
$t = \dfrac{{40}}{{15.28}}$
$\therefore t = 2.6\,s$
Which is enough time to enter the intersection before the light turns.

(b) In this case the total distance to stop (found in part(a) to be 34 m) is greater than the distance to the distance to the intersection(32 m), so the driver cannot stop without the front end of the car being a couple of meters into the intersection.
And the time to reach at constant speed it
$t = \dfrac{{32}}{{15.28}}$
$ \Rightarrow t = 2.1s$
Here $2.1s > 1.8\,s$ which is too long. The driver was caught in a rock and hard place.

Note: Force is the product of mass and acceleration. Application acceleration improves the application performances using techniques like compression, caching and transmission control protocol. When the velocity of any moving is changed it is said to be a change in acceleration. Change in acceleration may be positive or negative.