Answer to Question #236322 in Physics for eshiii

Question

2. A car starts from rest and accelerates 6 m/s2 for 5 s, after which it travels with a constant velocity for 9 s. The brakes are then applied so that it decelerates at 4 m/s2 . Find the total distance traveled by car.

Answer 1

Let's divide total distance $d$ into three parts:

1. $d_1$ with accelration $a_1 = 6m/s^2$ travelled during $t_1 = 5s$

2. $d_2$ with constant speed $v_2$ travelled during $t_2 = 9s$

3. $d_3$ with accelration $a_3 = -4m/s^2$.

The distance $d_1$ is given by the following kinematic equation

$d_1 = \dfrac{a_1t_1^2}{2}$

The speed of the car at the end of this distance (wich will be the constant speed for the second part) can be found from the definition of acceleration:

$v_2 = a_1t_1$

The distance $d_2$ travelled with the constant speed is given as follows:

$d_2 = v_2t_2 = a_1t_1t_2$

The distance $d_3$ can be found from another cinematic equation:

$d_3 = \dfrac{0^2-v_2^2}{2a_3} = -\dfrac{v_2^2}{2a_3} = -\dfrac{(a_1t_1t_2)^2}{2a_3}$

where represents the final speed of the car at the end of $d_3$ (car stops).

Thus, the total distance is:

$d = d_1 + d_2 + d_3\\ d = \dfrac{a_1t_1^2}{2} + v_2t_2 - -\dfrac{(a_1t_1t_2)^2}{2a_3}\\ d = \dfrac{6\cdot 5^2}{2} + 6\cdot 5\cdot 9 - \dfrac{(6\cdot 5\cdot 9) ^2}{-2\cdot 4} \approx 9458m$

Answer. 9458m.