Question

1. the distance the car traveled in meters at 2 sec, 4 sec, 6 sec and 8 sec were recorded in the data table below. calculate the average velocity for the car after 4 seconds.

time (sec) | distance (m)
2 sec | 1 m
4 sec | 2 m
6 sec | 3 m
8 sec | 4 m
solution

Explanation:

Step1: Recall the formula for average velocity

The formula for average velocity \( v_{avg} \) is \( v_{avg}=\frac{\Delta d}{\Delta t} \), where \( \Delta d \) is the change in distance and \( \Delta t \) is the change in time. We need to find the average velocity after 4 seconds, so we can consider the time interval starting from 4 seconds. Let's take the time from 4 sec to 8 sec (or any interval after 4 sec, but let's use the given data). From the table, at \( t_1 = 4 \) sec, \( d_1=2 \) m; at \( t_2 = 8 \) sec, \( d_2 = 4 \) m.

Step2: Calculate the change in distance and time

First, calculate \( \Delta d=d_2 - d_1=4 - 2 = 2 \) m. Then, calculate \( \Delta t=t_2 - t_1=8 - 4 = 4 \) sec.

Step3: Calculate the average velocity

Using the formula \( v_{avg}=\frac{\Delta d}{\Delta t} \), substitute \( \Delta d = 2 \) m and \( \Delta t = 4 \) sec. So \( v_{avg}=\frac{2}{4}=0.5 \) m/s. Alternatively, we can check with a smaller interval, say from 4 sec to 6 sec: \( \Delta d=3 - 2 = 1 \) m, \( \Delta t=6 - 4 = 2 \) sec, \( v_{avg}=\frac{1}{2}=0.5 \) m/s. Or from 6 sec to 8 sec: \( \Delta d=4 - 3 = 1 \) m, \( \Delta t=8 - 6 = 2 \) sec, \( v_{avg}=\frac{1}{2}=0.5 \) m/s. So the average velocity after 4 seconds is constant (uniform motion) and is 0.5 m/s.

Answer:

The average velocity for the car after 4 seconds is \( 0.5 \) m/s.