Question

A car reaches a final velocity of 22m/s, accelerating at 8 m/s^2 North from rest. How far did the car travel?

Answer 1

The car traveled a distance of 30.25 meters.

Step-by-step explanation:

To find the distance traveled by the car, we can use the equation of motion:

v^2 = u^2 + 2as

where:

- v is the final velocity (22 m/s in this case)

- u is the initial velocity (0 m/s since the car starts from rest)

- a is the acceleration (8 m/s^2 North in this case)

- s is the distance traveled (what we want to find)

Plugging in the given values into the equation, we get:

(22 m/s)^2 = (0 m/s)^2 + 2 * 8 m/s^2 * s

Simplifying, we have:

484 m^2/s^2 = 0 m^2/s^2 + 16 m/s^2 * s

Now we can solve for s:

484 m^2/s^2 = 16 m/s^2 * s

Dividing both sides of the equation by 16 m/s^2:

s = 484 m^2/s^2 / 16 m/s^2

s = 30.25 m^2

Therefore, the car traveled a distance of 30.25 meters.

Answer 2

The car traveled 30.25 meters.

Step-by-step explanation:

To calculate the distance traveled by the car, we can use the following formula, which relates the final velocity (v_final), initial velocity (v_initial), acceleration (a), and distance (d) of an object in motion:

d = (v_final^2 - v_initial^2) / (2a)

Plugging in the given values, we get:

d = (22^2 - 0^2) / (2 * 8)

d = 484 / 16

d = 30.25 meters

Since we are dealing with physical measurements, we round the distance to the nearest meter, which is 30 meters. Therefore, the car traveled 30.25 meters.