Question

Superman is flying 54.5 m/s when he sees

Answer 1

348.34 m/s. When Superman reaches the train, his final velocity will be 348.34 m/s.

To solve this problem, we are going to use the kinematics equations for constant aceleration. The key for this problem are the equations $d=v_{0} t+\frac{at^{2} }{2}$ and $v_{f} =v_{0} +at$ where $d$ is distance, $v_{0}$ is the initial velocity, $v_{f}$ is the final velocity, $t$ is time, and $a$ is aceleration.

Superman's initial velocity is $v_{0}=54.5\frac{m}{s}$, and he will have to cover a distance d = 850m in a time t = 4.22s. Since we know $d$, $v_{0}$ and $t$, we have to find the aceleration $a$ in order to find $v_{f}$.

From the equation $d=v_{0} t+\frac{at^{2} }{2}$ we have to clear $a$, getting the equation as follows: $a=\frac{2(d-v_{0}t) }{t^{2} }$.

Substituting the values:

$a=\frac{2(850m-54.5\frac{m}{s}.4.22s) }{(4.22s)^{2}}=69.63\frac{m}{s^{2}}$

To find $v_{f}$ we use the equation $v_{f} =v_{0} +at$.

Substituting the values:

$v_{f} =54.5\frac{m}{s} +(69.63\frac{m}{s^{2}}.4.22s)=348.34\frac{m}{s}$