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

Answer:

x = 265.5 m

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:

$x=\frac{v^2-v_o^2}{2a}\\\\x=\frac{(110m/s)^2-(65.0m/s)^2}{2(15.0m/s^2)}=262.5\ m$

hence, the distance traveled by the car is 265.5 m