Question

A car moving in a straight line starts at x = 0 at t = 0 . It passes the point x = 30.0 m with a speed of 10.0 m/s at t = 3.00 s . It passes the point x = 375 m with a speed of 50.0 m/s at t = 20.0 s . Find the average velocity between t = 3.00 s and t = 20.0 s . Express your answer with the appropriate units.
Find the average acceleration between t = 3.00 s and t = 20.0 s .
Express your answer with the appropriate units.

Answer 1

Answer:

a) $v=20.3m/s$

b) $a=2.35m/s^2$

Explanation:

From the exercise we know:

$x_{1}=30m$; $v_{1}=10m/s$; $t_{1}=3s$

$x_{2}=375m$; $v_{2}=50m/s$; $t_{2}=20s$

The formula for average velocity is:

$v=\frac{x_{2}-x_{1}}{t_{2}-t_{1}}$

a) $v=\frac{375m-30m}{20s-3s}=20.3m/s$

The formula for average acceleration is:

$a=\frac{v_{2}-v_{1}}{t_{2}-t_{1}}$

b) $a=\frac{(50-10)m/s}{(20-3)s}=2.35m/s^2$