Solve the problem.
-A rocket lifts off the surface of Earth with a constant acceleration of $30 \mathrm {~m} / \mathrm { sec } ^ { 2 }$. How fast will the rocket be going $2.5$ minutes later?
A) $3375 \mathrm {~m} / \mathrm { sec }$
B) $4500 \mathrm {~m} / \mathrm { sec }$
C) $- 4500 \mathrm {~m} / \mathrm { sec }$
D) $2250 \mathrm {~m} / \mathrm { sec }$

Question

Quizplus · Accepted Answer

We can use the equation of motion: 
$$v = u + at$$
Where:
$v$ = final velocity (to be found)
$u$ = initial velocity = 0 (since the rocket starts from rest)
$a$ = acceleration = $30~\mathrm{m}/\mathrm{sec}^2$
$t$ = time = $2.5$ minutes = $150$ seconds

Substituting the given values, we get: 
$$v = 0 + 30 \times 150 = 4500 ~\mathrm{m}/\mathrm{sec}$$

Therefore, the rocket will be going at a speed of $\textbf{4500 m/sec}$ after $2.5$ minutes.

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