Question

24 over a period of 2.50 seconds, a speedboat accelerates uniformly from 18.5 m/s to 46.5 m/s. what is the acceleration of the speedboat relative to the shore? f -28 m/s² g 28 m/s² h -11.2 m/s² j 11.2 m/s²

Explanation:

Step1: Recall the acceleration formula

The formula for acceleration \( a \) is \( a=\frac{v_f - v_i}{t} \), where \( v_f \) is the final velocity, \( v_i \) is the initial velocity, and \( t \) is the time.

Step2: Identify the given values

Here, \( v_i = 18.5\space m/s \), \( v_f = 46.5\space m/s \), and \( t = 2.50\space s \).

Step3: Substitute the values into the formula

Substitute the values into \( a=\frac{v_f - v_i}{t} \):
\( a=\frac{46.5 - 18.5}{2.50} \)

Step4: Calculate the numerator

First, calculate the numerator: \( 46.5 - 18.5 = 28 \)

Step5: Calculate the acceleration

Then, divide by the time: \( a=\frac{28}{2.50}=11.2\space m/s^2 \)

Answer:

J \( 11.2\space m/s^2 \)