Chuck’s car is moving at 65.0m/s when he suddenly accelerates his car at 15.0m/s2 for 3.00s. How far did Chuck, and his car, travel while he was accelerating?

Answers

Answer:

x = 265.5 m

Explanation:

To find the distance traveled by the car you first use the following formula:

(1)

vo: initial velocity = 65.0m/s

a: acceleration = 15.0m/s^2

t: time = 3.00 s

you replace the values of vo, t and a in the equation (1) in order to calculate the final velocity v:

= 110 m/s

Next, you use the following formula:

x: distance traveled

You do x the subject of the formula and replace the values of vo, v and a:

hence, the distance traveled by the car is 265.5 m

Answer:x = 265.5 mExplanation:To find the distance traveled by the car you first use the following formula:

(1)

vo: initial velocity = 65.0m/s

a: acceleration = 15.0m/s^2

t: time = 3.00 s

you replace the values of vo, t and a in the equation (1) in order to calculate the final velocity v:

= 110 m/s

Next, you use the following formula:

x: distance traveled

You do x the subject of the formula and replace the values of vo, v and a:

hence, the distance traveled by the car is 265.5 m