MATH SOLVE

2 months ago

Q:
# A rock dropped on the moon will increase its speed from 0 m/s (its startingspeed) to 8.15 m/s in about 5 seconds. If you were standing on the moon,measuring the rock's motion, what would you measure for the magnitude ofits acceleration?OA. 8.2 m/s2O B. 41 m/s2O C. 9.8 m/s2O D. 1.63 m/s2

Accepted Solution

A:

Hello!The answer is:The correct option is the option D[tex]g=1.63\frac{m}{s^{2}}[/tex]Why?To calculate the acceleration, we need to use the following formula that involves the given information (initial speed, final speed, and time).We need to use the following free fall equation:[tex]V_{f}=V_{o}-g*t[/tex]Where:[tex]V_{f}[/tex] is the final speed.[tex]V_{o}[/tex] is the initial speed.g is the acceleration due to gravity.t is the time.We are given the following information:[tex]V_{f}=8.15\frac{m}{s}[/tex][tex]V_{o}=0\frac{m}{s}[/tex][tex]t=5seconds[/tex]Then, using the formula to isolate the acceleration, we have:[tex]V_{f}=V_{o}-g*t[/tex][tex]V_{f}=V_{o}-g*t\\\\g*t=V_{o}-V_{f}\\\\g=\frac{V_{o}-V_{f}}{t}[/tex]Now, substituting we have:[tex]g=\frac{V_{o}-V_{f}}{t}[/tex][tex]g=\frac{0-8.15\frac{m}{s}}{5seconds}=-1.63\frac{m}{s^{2}}[/tex]Therefore, since we are looking for a magnitude, we have that the obtained value will be positive, so:[tex]g=1.63\frac{m}{s^{2}}[/tex]Hence, the correct option is the option D[tex]g=1.63\frac{m}{s^{2}}[/tex]Have a nice day!