Question

South African frogs are capable of jumping as far as 10.0m in one hop.Suppose one of these frogs makes exactly 15 of these jumps in a time interval of 60.0s.

a. What is the frog's average velocity?
b. If the frog lands with a velocity equal to it average velocity and comes to a full stop 0.25 s later, what is the frog's average acceleration?

Answer 1

The frog's average velocity is 2.5 m/s. The average acceleration as the frog comes to a stop is -10 m/s², which indicates deceleration.

Step-by-step explanation:

To calculate the average velocity of a South African frog that makes 15 jumps of 10.0 m each in 60.0 s, we use the formula for average velocity, which is the total displacement divided by the total time. The total displacement is 15 jumps × 10.0 m/jump = 150 m. The time is 60.0 s. Thus, the average velocity is 150 m / 60.0 s = 2.5 m/s.

To determine the frog's average acceleration upon landing, we use the formula a = Δv / Δt, where Δv is the change in velocity and Δt is the change in time. Since the frog comes to a stop, the final velocity is 0 m/s, and the initial velocity is the average velocity (2.5 m/s). Thus, the change in velocity is -2.5 m/s because the frog is decelerating. The time taken to stop is 0.25 s. Therefore, the average acceleration is -2.5 m/s / 0.25 s = -10 m/s². The negative sign indicates that the acceleration is in the opposite direction of the velocity (deceleration).