A mouse is running away from a cat at 0.75 m/s and,at 1.25 m from his mouse hole,he accelerates at a constant rate until he reaches his hole safely,0.75 s later.What acceleration did he undergo during the final 1.25 m of his escape? A) 1.8 m/s² B) 1.2 m/s² C) 4.8 m/s² D) 2.4 m/s² E) 6.4 m/s²

Question

Quizplus · Accepted Answer

We can use the following equation to solve the problem:

Δx = ut + 0.5at^2

where:

Δx = displacement (1.25 m) 
u = initial velocity (0.75 m/s) 
t = time (0.75 s) 
a = acceleration (unknown)

Rearranging the equation:

a = (2Δx - 2ut) / t^2

Substituting the values:

a = (2(1.25) - 2(0.75)(0.75)) / (0.75)^2

a = 2.4 m/s^2

Therefore, the mouse underwent an acceleration of 2.4 m/s^2 during the final 1.25 m of its escape.

A Mouse Is Running Away from a Cat at 0