Question

A car movingin a straight line starts at x = 0 at t = 0. It passes the point x = 25.0 m with a speed of 11.0 m????s at t = 3.00s. It passes the point x = 385m with a speed of 45.0 m/s at t = 20.0 s. Find (a) the average velocity, and (b) the average acceleration, between t = 3.00 s and t = 20.0 s.

Answer 1

Answer:

Part a)

at t = 3.00 s

$v = 8.33 m/s$

at t = 20.0 s

$v = 19.25 m/s$

Part b)

at t = 3.00 s

$a = 3.67 m/s^2$

at t = 20.0 s

$a = 2.25 m/s^2$

Explanation:

The car starts at x = 0

Part a)

Now at t = 3.00 s

the position of the car is given as x = 25 m and its speed is given as v = 11 m/s

Now for average velocity we have

$v = \frac{displacement}{time}$

$v = \frac{25 - 0}{3}$

$v = 8.33 m/s$

Now for average acceleration we have

$a = \frac{v - 0}{t}$

$a = \frac{11 - 0}{3}$

$a = 3.67 m/s^2$

Part b)

Now at t = 20.0 s

the position of the car is given as x = 385 m and its speed is given as v = 45 m/s

Now for average velocity we have

$v = \frac{displacement}{time}$

$v = \frac{385 - 0}{20}$

$v = 19.25 m/s$

Now for average acceleration we have

$a = \frac{v - 0}{t}$

$a = \frac{45 - 0}{20}$

$a = 2.25 m/s^2$