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 these two phases is 51.9 m when its initial speed is 78.3 km/h, and 23.6 m when its initial speed is 49.1 km/h. What are (a) your reaction time and (b) the magnitude of the deceleration?

Answer 1

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 these two phases is $51.9\mathrm{m}$ when its initial speed is $78.3\mathrm{km / h}$ , and $23.6\mathrm{m}$ when its initial speed is $49.1\mathrm{km / h}$ . What is (a) your reaction time?

Solution.

For both cases (initial velocities $v_{1}$ and $v_{2}$ ) we right down Newton's equations:

$$d ^ {2} / t ^ {2} x = 0, 0 < t < t _ {0},$$ $$d ^ {2} / t ^ {2} x = - a _ {s}, t _ {0} < t,$$

where $t_0$ is the reaction time and $a_s$ is the deceleration value. We solve these equations for different velocity values:

$$x = v _ {1} t, 0 < t < t _ {0},$$ $$x = - a t ^ {2} / 2 + v _ {1} t, t _ {0} < t$$

for $v = v_{2}$

$$x = v _ {2} t, 0 < t < t _ {0},$$ $$x = - a t ^ {2} / 2 + v _ {2} t, t _ {0} < t$$

Since reaction time $t_0$ in both cases is the same, we introduce the length of the path before driver's reaction

$$l _ {1, 2} = v _ {1, 2} t _ {0}.$$

Deceleration continues until car stops. We find duration of deceleration requiring $dx/dt = v(t_{d1,2}) = 0$

$$v \left(t _ {d 1, 2}\right) = - a \left(t _ {d 1, 2}\right) + v _ {1, 2} = 0$$ $$t _ {d 1, 2} = v _ {1, 2} / a$$

Putting it into the main equation, we obtain

$$d _ {1, 2} = v _ {1, 2} t _ {0} + v \frac {v _ {1 , 2}}{a} - \frac {a}{2} \left(\frac {v _ {1 , 2}}{a}\right) ^ {2}$$ $$d _ {1, 2} = v _ {1, 2} t _ {0} + \frac {v _ {1 , 2} ^ {2}}{2 a}$$ $$\frac {1}{2} \frac {v _ {1 , 2} ^ {2}}{d _ {1 , 2} - v _ {1 , 2} t _ {0}} = a$$ $$\frac {v _ {1} ^ {2}}{d _ {1} - v _ {1} t _ {0}} = \frac {v _ {2} ^ {2}}{d _ {2} - v _ {2} t _ {0}}$$ $$\frac {d _ {1} - v _ {1} t _ {0}}{v _ {1} ^ {2}} = \frac {d _ {2} - v _ {2} t _ {0}}{v _ {2} ^ {2}}$$ $$t _ {0} = \frac {v _ {1} ^ {2} d _ {2} - v _ {2} ^ {2} d _ {1}}{v _ {1} v _ {2} ^ {2} - v _ {2} v _ {1} ^ {2}} = \frac {(7 8 . 3 k m / h) ^ {2} 2 3 . 6 m - (4 9 . 1 k m / h) ^ {2} 5 9 . 9 m}{(7 8 . 3 k m / h) (4 9 . 1 k m / h) ^ {2} - (4 9 . 1 k m / h) (7 8 . 3 k m / h) ^ {2}}$$ $$t _ {0} = \frac {(2 1 . 7 5 m / s) ^ {2} 2 3 . 6 m - (1 3 . 6 4 m / s) ^ {2} 5 9 . 9 m}{(2 1 . 7 5 m / s) (1 3 . 6 4 m / s) ^ {2} - (1 3 . 6 4 m / s) (2 1 . 7 5 m / s) ^ {2}} = - 0. 0 0 8 s$$

Answer.

$$t _ {0} = - 0. 0 0 8 s$$

What is (b) the magnitude of the deceleration?

Solution.

$$a = \frac {1}{2} \frac {v _ {1 , 2} ^ {2}}{d _ {1 , 2} - v _ {1 , 2} t _ {0}} = 0. 5 \frac {(2 1 . 7 5 m / s) ^ {2}}{2 3 . 6 m - 2 1 . 7 5 m / s * 0 . 0 0 8 s} = 1 0. 1 m / s ^ {2}$$

Answer.

$$a = 1 0. 1 m / s ^ {2}$$