Answer to Question #181613 in Classical Mechanics for kan

Question

a car accelerate from rest to reach a certain speed in 10 min. it then continues at this speed for another 10 minutes and decelerate to rest in another 5 min, the total distance covered is 17.5km. find the steady speed reached

Answer 1

Let the times of motion during three different parts of the distance be $t_1, t_2, t_3$, the constant speed after the acceleration $v_1$, and accelerations during first and third intervals be $a_1, a_3$. Then the corresponding displacements and speeds during three parts of the way are:

1) $s_1 = \frac{a_1 t_1^2}{2}$, $v_1 = a_1 t_1$

2) $s_2 = v_1 t_2$

3) $s_3 = v_1 t_3 - \frac{a_3 t_3^2}{2}, 0 = v_1 - a_3 t_3$.

Total distance is $s = s_1 + s_2 + s_3 = \frac{a_1 t_1^2}{2} + v_1 (t_2 + t_3) - \frac{a_3 t_3^2}{2}$.

From 1) and 3), $v_1 = a_1 t_1 = a_3 t_3$, from where $a_3 = \frac{t_1}{t_3} a_1$. Substituting the last expression for $a_3$ and $v_1 = a_1 t_1$ from 1) into $s$, obtain:

$s = \frac{a_1 t_1^2}{2} + a_1 t_1(t_2 + t_3) - \frac{a_1 t_1 t_3}{2}$, from where $a_1 = \frac{s}{\frac{t_1^2}{2} + t_1 t_2 + \frac{t_1 t_3}{2}} = 0.1 \frac{km}{min^2}$.

The steady speed is hence $v_1 = a_1 t_1 = 0.1 \frac{km}{min^2} \cdot 10 min = 1\frac{km}{min}$.