Question

in 45 seconds, a rocket goes from traveling at 75 m/s to 570 m/s. what is the acceleration of the rocket? a = ? $\frac{m}{s^{2}}$ enter the sign (+ or -) and the magnitude. remember: a = $\frac{delta v}{t}$

Explanation:

Step1: Identify initial and final velocities and time

Initial velocity $v_0 = 75$ m/s, final velocity $v = 570$ m/s, time $t=45$ s.

Step2: Calculate the change in velocity

$\Delta v=v - v_0=570 - 75=495$ m/s.

Step3: Calculate acceleration

Using the formula $a=\frac{\Delta v}{t}$, we substitute $\Delta v = 495$ m/s and $t = 45$ s. So $a=\frac{495}{45}=11$ m/s². Since the velocity is increasing, the sign is positive.

Answer:

$+ 11$ m/s²