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

The total time elapsed from start to stop for the electric vehicle is the sum of the time taken to accelerate and decelerate, calculated as T = v / a1 + v / a2. The total distance moved is the sum of the distances during acceleration and deceleration given by S = 0.5 * a1 * (v / a1)^2 + 0.5 * a2 * (v / a2)^2.

Step-by-step explanation:

The question inquires about the time elapsed and the distance traveled by an electric vehicle which accelerates from rest until it reaches a certain velocity, and then decelerates to a stop. To answer part (a), we need to calculate the time taken for both acceleration and deceleration phases. For acceleration, we use the formula t1 = v / a1, and for deceleration t2 = v / a2. The total time elapsed T is then the sum of t1 and t2.

For part (b), the total distance covered S is the sum of the distances covered during acceleration and deceleration. The distance covered during acceleration s1 can be found using the equation s1 = 0.5 * a1 * t1^2, and the distance covered during deceleration s2 can be found with s2 = 0.5 * a2 * t2^2.

Hence, for an electric vehicle that accelerates to a speed v at a rate a1 and then decelerates at a rate a2 to a stop:

• The total time elapsed is T = t1 + t2, which simplifies to T = v / a1 + v / a2.
• The total distance moved is S = s1 + s2, which simplifies to S = 0.5 * a1 * (v / a1)^2 + 0.5 * a2 * (v / a2)^2.

Answer 2

(a)
`t=(v)/(a_1)+(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=(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=(v)/(a_2)` (2)

So the total time taken for the trip is

`t=(v)/(a_1)+(v)/(a_2)`

(b)
`d=(v^2)/(2a_1)+(v^2)/(2a_2)`

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

`d_1 = u t_1 + (1)/(2)a_1 t_1^2`

Substituting u = 0 and
`t_1=(v)/(a_1)` (1), we find

`d_1 = (1)/(2)a_1 (v^2)/(a_1^2) = (v^2)/(2a_1)`

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

`d_2 = v t_2 - (1)/(2)a_2 t_2^2`

Substituting
`t_2=(v)/(a_2)` (2), we find

`d_1 = (v^2)/(a_2) - (1)/(2) (v^2)/(a_2) = (v^2)/(2a_2)`

So the total distance travelled is

`d= d_1 +d_2 = (v^2)/(2a_1)+(v^2)/(2a_2)`