Motorcycle's Velocity
A motorcycle is accelerating at a constant rate of 10 m/sec^2. It travels 120 meters while its speed changes to 50 m/s. What is the initial velocity?
Hint
$$$v=v_0+at$$$
where
$$v$$
is the velocity,
$$v_0$$
is the initial velocity,
$$a$$
is acceleration, and
$$t$$
is time.
Hint 2
$$$x=x_0+v_0t+\frac12at^2$$$
where
$$x$$
is the distance,
$$x_0$$
is the initial distance,
$$v_0$$
is the initial velocity,
$$t$$
is the time, and
$$a$$
is the acceleration.
The below equations relate displacement, velocity, acceleration, and time, and apply for constant acceleration in the +x direction starting at t=0.
$$$v=v_0+at\:\:\:\:\:\:\:\:\:\:(1)$$$
where
$$v$$
is the velocity,
$$v_0$$
is the initial velocity,
$$a$$
is acceleration, and
$$t$$
is time.
$$$x=x_0+v_0t+\frac12at^2\:\:\:\:\:\:\:\:\:\:(2)$$$
where
$$x$$
is the distance,
$$x_0$$
is the initial distance,
$$v_0$$
is the initial velocity,
$$t$$
is the time, and
$$a$$
is the acceleration.
Using Eq. (1),
$$a=(v-v_0)/t$$
, in Eq. (2), we get:
$$$x=x_0+v_0t+\frac12(v-v_0)t$$$
$$$=x_0+\frac12(v+v_0)t\:\:\:\:\:\:\:\:\:\:(3)$$$
$$$=x_0+v_{avg}t$$$
Rearranging (1),
$$t=(v-v_0)/a$$
, and substituting for
$$t$$
in (3) gives:
$$$x=x_0+\frac12(v+v_0)(v-v_0)/a$$$
$$$2a(x-x_0)=v^2-v_0^2\:\:\:\:\:\:\:\:\:\:(4)$$$
$$$v_0^{2}=v^{2}-2a(x-x_0)=(50\frac{m}{s})^2-2(10\frac{m}{s^{2}})(120m)=100(\frac{m}{s})^2$$$
$$$v_o=10\frac{m}{s}$$$
10 m/s