Question

a car accelerates from rest at for 50 m, then brakes with an acceleration of and comes to a stop. what was the total distance traveled in m? a car accelerates from rest at for 50 m, then brakes with an acceleration of and comes to a stop. what was the total distance traveled in m? 85 90 95 100 105

Answer 1

Based on the given information, the total distance traveled by the car is approximately 95 m. C

What was the total distance traveled?

Let's break down the motion of the car into two parts: the period of acceleration and the period of braking.

Acceleration Phase:

The car starts from rest and accelerates at 8.3m/s^2 for a distance of 50m.

We can use the equation of motion:

`v^2 =u^2 +2as`

where:

v = final velocity

u = initial velocity (which is 0 as the car starts from rest)

a = acceleration

s = distance

We need to find the final velocity after accelerating for a distance of 50m:

`v^2 =0^2 + 2 * 8.3 m/s^2 * 50\\v^2 = 2 * 8.3 * 50\\v^2 = 830\\v = \sqrt830`

v ≈28.8m/s

So, after the acceleration phase, the car reaches a velocity of approximately 28.8m/s.

Braking Phase:

Next, the car brakes with an acceleration of 9.2m/
`s^2` until it comes to a stop. To find the distance traveled during braking, we can use the equation of motion again:

`v^2 =u^2 +2as`

where:

v = final velocity (which is 0 as the car stops)

u = initial velocity (which is 28.8 m/s after acceleration)

a = acceleration (braking, which is 9.2m/
`s^2` )

s = distance

`0 = 28.8^2 + 3 * (-9.2) * s`

0 = 829.44 - 88.4s

s = 829.44/88.4 = 45 m

Therefore, during the braking phase, the car traveled approximately

The total distance traveled by the car is the sum of the distances covered during acceleration and braking:

Total distance =50 m + 45 m = 95 m.

So, the total distance traveled by the car is approximately 95m.

A car accelerates from rest at 8.3m/s^2 for 50 m, then brakes with an acceleration of 9.2m/s^2 and comes to a stop. What was the total distance traveled in m? 85 90 95 100 105