## 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