Question

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?

Answer 1

x = 265.5 m

Step-by-step explanation:

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

`v=v_o+at` (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:

`v=65.0m/s+15.0m/s^2(3.00s)` = 110 m/s

Next, you use the following formula:

`v^2=v_o^2+2ax`

x: distance traveled

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

`x=(v^2-v_o^2)/(2a)\\\\x=((110m/s)^2-(65.0m/s)^2)/(2(15.0m/s^2))=262.5\ m`

hence, the distance traveled by the car is 265.5 m