Question

An electric vehicle starts from rest and accelerates at a rate a1 in a straight line until it reaches a speed of v. The vehicle then slows at a constant rate a2 until it stops. (a) How much time elapses from start to stop? (b) How far does the vehicle move from start to stop? Give your answers in terms of the given variables.

Answer 1

(a) $t=\frac{v}{a_1}+\frac{v}{a_2}$

In the first part of the motion, the car accelerates at rate $a_1$, so the final velocity after a time t is:

$v = u +a_1t$

Since it starts from rest,

u = 0

So the previous equation is

$v= a_1 t$

So the time taken for this part of the motion is

$t_1=\frac{v}{a_1}$ (1)

In the second part of the motion, the car decelerates at rate $a_2$, until it reaches a final velocity of v2 = 0. The equation for the velocity is now

$v_2 = v - a_2 t$

where v is the final velocity of the first part of the motion.

Re-arranging the equation,

$t_2=\frac{v}{a_2}$ (2)

So the total time taken for the trip is

$t=\frac{v}{a_1}+\frac{v}{a_2}$

(b) $d=\frac{v^2}{2a_1}+\frac{v^2}{2a_2}$

In the first part of the motion, the distance travelled by the car is

$d_1 = u t_1 + \frac{1}{2}a_1 t_1^2$

Substituting u = 0 and $t_1=\frac{v}{a_1}$ (1), we find

$d_1 = \frac{1}{2}a_1 \frac{v^2}{a_1^2} = \frac{v^2}{2a_1}$

In the second part of the motion, the distance travelled is

$d_2 = v t_2 - \frac{1}{2}a_2 t_2^2$

Substituting $t_2=\frac{v}{a_2}$ (2), we find

$d_1 = \frac{v^2}{a_2} - \frac{1}{2} \frac{v^2}{a_2} = \frac{v^2}{2a_2}$

So the total distance travelled is

$d= d_1 +d_2 = \frac{v^2}{2a_1}+\frac{v^2}{2a_2}$