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4 years ago

A city bus travels along its route from Jackson Street 8

Physics

1 answer:

wel4 years ago

6 0

Answer:21km

Explanation:13km+8km=21km

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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

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(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

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

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