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) $37.5 \mathrm {~m} / \mathrm { sec }$
B) $187.5 \mathrm {~m} / \mathrm { sec }$
C) $- 75 \mathrm {~m} / \mathrm { sec }$
D) $75 \mathrm {~m} / \mathrm { sec }$

Question

Quizplus · Accepted Answer

We can use the equation for velocity with constant acceleration:
$$v = v_0 + at$$
where
$v$ = final velocity
$v_0$ = initial velocity (in this case, 0)
$a$ = acceleration (given as $30 \mathrm {~m} / \mathrm { sec } ^ { 2 }$)
$t$ = time (given as $2.5$ minutes or $150$ seconds)

Substituting the given values, we get:
$$v = 0 + (30 \mathrm {~m} / \mathrm { sec } ^ { 2 })(150 \mathrm {~sec})$$
$$v = 4500 \mathrm {~m} / \mathrm { sec }$$
Converting to meters per second:
$$v = 4500 \mathrm {~m} / \mathrm { sec } \div 60 \mathrm {~sec} / \mathrm {min} = 75 \mathrm {~m} / \mathrm { sec }$$

Therefore, the rocket will be going at a speed of $75 \mathrm {~m} / \mathrm { sec }$ after $2.5$ minutes. The answer is D.

Solve the Problem 30 m/sec230 \mathrm {~m} / \mathrm { sec } ^ { 2 }30 m/sec2

Solve the Problem $30 \mathrm {~m} / \mathrm { sec } ^ { 2 }$