Question

1. 8a sports car approaches a highway on - ramp at a velocity of 20.0 m/s e. if the car accelerates at a rate of 3.2 m/s² e for 5.0 s, what is the displacement of the car?

Explanation:

Step1: Identify the kinematic equation

We use the kinematic equation for displacement with constant acceleration: \( s = v_0t+\frac{1}{2}at^2 \), where \( v_0 \) is the initial velocity, \( t \) is time, and \( a \) is acceleration.

Step2: Substitute the given values

Given \( v_0 = 20.0\ m/s \), \( a = 3.2\ m/s^2 \), \( t = 5.0\ s \).
Substitute into the equation: \( s=(20.0)(5.0)+\frac{1}{2}(3.2)(5.0)^2 \)

Step3: Calculate each term

First term: \( (20.0)(5.0)=100\ m \)
Second term: \( \frac{1}{2}(3.2)(25)=40\ m \)

Step4: Sum the terms

\( s = 100 + 40 = 140\ m \) (direction is [E], but we are asked for displacement magnitude here)

Answer:

The displacement of the car is \( 140\ m \) [E] (or just \( 140\ m \) considering the magnitude with the given direction context).