Question

car travels between 2 stations .60 km apart. Left first station, accelerates for 10.4 s at 1 m/s^2 and travels @ a constant speed until nearing the second station, when it brakes at 2.0 m/s^2 in order to stop at the station. How long did this trip take

Answer 1

Question #23588

car travels between 2 stations .60 km apart. Left first station, accelerates for 10.4 s at 1 m/s^2 and travels @ a constant speed until nearing the second station, when it brakes at 2.0 m/s^2 in order to stop at the station. How long did this trip take

Solution:

Let:

$$S = 0.60 \text{ km} = 600 \text{ m}$$ $$t_1 = 10.4 \text{ s}$$ $$a_1 = 1 \text{ m/s}^2$$ $$a_2 = 2 \text{ m/s}^2$$ $$t = ?$$ $$t = t_1 + t_c + t_2,$$

were $t_c$ is the time of traveling with a constant speed, $t_2$ is the time of braking

$$t_c = \frac{S_c}{v}$$

were $S_c$ is the distance of traveling with a constant speed, $v$ is the velocity

$$v = a_1 t_1$$ $$t_2 = \frac{v}{a_2} = \frac{a_1 t_1}{a_2}$$ $$S_c = S - (S_1 + S_2)$$

were $S_1$ is the distance of traveling with acceleration, $S_2$ is the distance of braking

$$S_1 = \frac{1}{2} a_1 t_1^2$$ $$S_2 = \frac{1}{2} a_2 t_2^2 = \frac{1}{2} a_2 \left(\frac{a_1 t_1}{a_2}\right)^2 = \frac{a_1^2 t_1^2}{2 a_2}$$ $$t_c = \frac{S - \left(\frac{1}{2} a_1 t_1^2 + \frac{a_1^2 t_1^2}{2 a_2}\right)}{a_1 t_1}$$ $$t = t_1 + \frac{S - \left(\frac{1}{2} a_1 t_1^2 + \frac{a_1^2 t_1^2}{2 a_2}\right)}{a_1 t_1} + \frac{a_1 t_1}{a_2} = 10.4 + \frac{600 - \left(\frac{1}{2} t_1 \cdot 10.4^2 + \frac{1^2 10.4^2}{2 \cdot 2}\right)}{1 \cdot 10.4} + \frac{1 \cdot 10.4}{2} = 65.5 \text{ s}$$

Answer: 65.5 s (or 1 m 5.5 s).