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

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`

Step-by-step 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 = (displacement)/(time)`

`v = (25 - 0)/(3)`

`v = 8.33 m/s`

Now for average acceleration we have

`a = (v - 0)/(t)`

`a = (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 = (displacement)/(time)`

`v = (385 - 0)/(20)`

`v = 19.25 m/s`

Now for average acceleration we have

`a = (v - 0)/(t)`

`a = (45 - 0)/(20)`

`a = 2.25 m/s^2`