Question

8/29/2021

1. a speedboat increases its speed uniformly from 20 m/s to 30 m/s in a distance of 200m. find the magnitude of its acceleration and the time it takes the boat to travel the 200 - m distance.

vi = 20 m/s
vf = 30 m/s
a = ?
d = 200 m
30² = 20² + 2a×200
900 = 400 + 400a
500 = 400a
a = 1.25 m/s²

Explanation:

Step1: Identify the kinematic equation

We use the kinematic equation \( v_f^2 = v_i^2 + 2ad \) to find acceleration, where \( v_i = 20 \, \text{m/s} \), \( v_f = 30 \, \text{m/s} \), \( d = 200 \, \text{m} \), and \( a \) is acceleration.
\[
30^2 = 20^2 + 2a(200)
\]

Step2: Solve for acceleration \( a \)

First, calculate the squares: \( 900 = 400 + 400a \). Subtract 400 from both sides: \( 500 = 400a \). Then divide by 400: \( a=\frac{500}{400}=1.25 \, \text{m/s}^2 \).

Step3: Use another kinematic equation for time

Use \( d=\frac{v_i + v_f}{2}t \) to find time \( t \). Substitute values: \( 200=\frac{20 + 30}{2}t \). Simplify the right side: \( 200 = 25t \). Solve for \( t \): \( t=\frac{200}{25}=8 \, \text{s} \).

Answer:

Acceleration: \( 1.25 \, \text{m/s}^2 \), Time: \( 8 \, \text{s} \)