Question

To stop a car, first you require a certain reaction time to begin braking; then the car slows under the constant braking deceleration. Suppose that the total distance moved by your car during those two phases is \[56.7m\] when it’s initial speed is \[80.5km\,h{{r}^{-1}}\] and \[24.4\,m\] when it’s speed is \[48.3\,km\,h{{r}^{-1}}\] . What are-
(a). Your reaction time
(b). magnitude of deceleration

Answer 1

Hint: As the car is transitioning from acceleration to deceleration, it travels a given distance for every speed. Using the given parameters, we can calculate time. At the last second the car starts decelerating. So using the time calculated we can determine the magnitude of deceleration.

Formula used:
 \[s=ut+\dfrac{1}{2}a{{t}^{2}}\]

Complete step-by-step answer:
Acceleration is defined as the change in velocity per unit time. Its SI unit is \[m{{s}^{-2}}\]
The time difference between applying breaks and the deceleration to start is the reaction time.
There are some equations which represent the different relations between initial velocity ( \[{{v}_{{}}}\] ), final velocity ( \[{{v}_{{}}}\] ), displacement ( \[s\] ), time ( \[t\] ) and acceleration ( \[a\] ). They are-
 \[v=u+at\] - (1)
 \[{{v}^{2}}={{u}^{2}}+2as\] - (2)
 \[s=ut+\dfrac{1}{2}a{{t}^{2}}\] - (3)
We are given,
 \[\begin{align}
  & {{s}_{1}}=56.7m \\
 & {{u}_{1}}=80.5km\,h{{r}^{-1}}=22.36m{{s}^{-1}} \\
\end{align}\]
 \[\begin{align}
  & {{s}_{2}}=24.4m \\
 & {{u}_{2}}=48.3km\,h{{r}^{-1}}=13.42m{{s}^{-1}} \\
\end{align}\]
 \[\]
Substituting the given values in eq (3), we get,
 \[56.7=(22.36)t-\dfrac{1}{2}a{{t}^{2}}\] - (4) (since we are decelerating, \[a=-a\] )
 \[24.4=(13.42)t-\dfrac{1}{2}a{{t}^{2}}\] - (5)
Subtracting eq (4) and eq (5), we get,
 \[\begin{align}
  & 32.3=(8.94)t \\
 & \Rightarrow t=\dfrac{32.3}{8.94}=3.61\operatorname{s} \\
\end{align}\]
Therefore the reaction time is \[3.61s\]
Substituting the value of t in eq (5), we get
 \[24.4=13.42\times 3.61-\dfrac{1}{2}a{{(3.61)}^{2}}\]
 \[24.4=48.44-\dfrac{1}{2}a(13.03)\]
 \[\begin{align}
  & \Rightarrow 24.04=6.51a \\
 & \therefore a=3.69m{{s}^{-2}} \\
\end{align}\]
At \[3.61s\] the car will start decelerating with the deceleration of \[3.69m{{s}^{-2}}\]
Therefore, (a). The reaction time is \[3.61s\]
(b) Magnitude of deceleration is \[3.69m{{s}^{-2}}\] .

Note: In the reaction time the car will transition from acceleration to deceleration and at the last second it will start decelerating. Carefully convert the units as required. The direction of acceleration is the same as the direction of motion. The magnitude of Deceleration is always negative and the final velocity is 0.