## Constant Acceleration

A car accelerates at a constant rate of 12 m/sec^2. The car travels 140 meters while its speed changes to 60 m/sec. What is the initial velocity?

##
__
__**Hint**

**Hint**

$$$v^{2}=v_{o}^{2}+2a(S-S_{o})$$$

where
$$v$$
is the velocity along the direction of travel,
$$v_o$$
is the velocity at time
$$t_0$$
,
$$a$$
is constant acceleration,
$$S$$
is the displacement at time
$$t$$
along the line of travel, and
$$S_o$$
is the displacement at time
$$t_0$$
.

##
__
__**Hint 2**

**Hint 2**

Solve for initial velocity,
$$v_o$$
.

For constant acceleration, the equation for velocity as a function of position:

$$$v^{2}=v_{o}^{2}+2a(S-S_{o})$$$

where
$$v$$
is the velocity along the direction of travel,
$$v_o$$
is the velocity at time
$$t_0$$
,
$$a$$
is constant acceleration,
$$S$$
is the displacement at time
$$t$$
along the line of travel, and
$$S_o$$
is the displacement at time
$$t_0$$
.

Solving for initial velocity:

$$$v_{o}^{2}=v^{2}-2a(S-S_{o})$$$

$$$v_{o}^{2}=(60\:m/sec)^{2}-2(12\:m/sec^2)(140\:m)$$$

$$$v_{o}^{2}=3,600\:m^2/sec^{2}-3,360\:m^2/sec^2$$$

$$$v_{o}=\sqrt{240\:m^2/sec^2}= 15.5\:m/sec$$$

15.5 m/s