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

Answer:

a) $v_{o}=4.98m/s$

b) $a=1.67m/s^{2}$

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 $x=x_{o}+v_{o}t+\frac{1}{2}at^{2}$

Now, we have two equations with two unknowns

$v_{o}=v-at$ (1)

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

$x=x_{o}+(v-at)t+\frac{1}{2}at^{2}$

$60=vt-at^{2}+\frac{1}{2}at^{2}$

$60=vt-\frac{1}{2}at^{2}$

Solving for a

b) $a=\frac{2(vt-60)}{t^{2} } =\frac{2((15)(6)-60)}{(6)^{2} }=1.67m/s^{2}$

Now, we can solve (1)

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