Question

A car traveling at a constant acceleration covers the distance between two points 180m apart in 6 seconds if the speed of the car at the second point is 45 m/s what is the acceleration of the car?

Answer 1

Answer:
`15 m/s^(2)`

Step-by-step explanation:

We have the following data:

`d=180 m` is the distance traveled by the car

`t= 6 s` is the time of travel

`u=45 m/s` is the car's final velocity

`v` is the unknown car's initial velocity

`a` is the car's constant acceleration

Now, we can solve this problem with the following equations of motion:

`u=v+at` (1)

`d=vt+(1)/(2)at^(2)` (2)

Isolating
`v` from (1):

`v=u-at` (3)

`v=45 m/s-(6 s)a` (4)

Substituting (4) in (2):

`d=(45 m/s-(6 s)a)t+(1)/(2)at^(2)` (5)

Solving and finding
`a`:

`180 m=(45 m/s)(6 s)-36 s^(2)a+(1)/(2)a(6 s)^(2)` (6)

Finally:

`a=15 m/s^(2)` This is the car's acceleration