Question

A body moving with constant acceleration covers the distance between two points 60.0 m apart in 6.0 seconds. Its velocity as it passes the second point is 15 m/s. (a) What is its velocity at the first point? (b) What is the acceleration?

Answer 1

a)
`v_(o)=4.98m/s`

b)
`a=1.67m/s^(2)`

Step-by-step explanation:

From the exercise we got final position, final velocity and how much time does it takes the body to cover the distance

`x=60m\\t=6s\\v=15m/s`

From the concept of moving objects we know the following equations:

`v=v_(o)+at` and
`x=x_(o)+v_(o)t+(1)/(2)at^(2)`

Now, we have two equations with two unknowns

`v_(o)=v-at` (1)

Now, we need to replace (1) in the other equation

`x=x_(o)+(v-at)t+(1)/(2)at^(2)`

`60=vt-at^(2)+(1)/(2)at^(2)`

`60=vt-(1)/(2)at^(2)`

Solving for a

b)
`a=(2(vt-60))/(t^(2) ) =(2((15)(6)-60))/((6)^(2) )=1.67m/s^(2)`

Now, we can solve (1)

a)
`v_(o)=v-at=15-1.67(6)=4.98m/s`