Question

A car, initially traveling east with a speed of 5.0 meters per second, is accelerated uniformly at 2.0 meters per second 2 east for 10 seconds along a straight line. What is the total distance traveled by the car during this 10-second interval

Answer 1

To determine the total distance traveled by the car, we can use the kinematic equations of motion. The car is initially moving at a speed of \( v_0 = 5.0 \, \text{m/s} \) and is experiencing a uniform acceleration \( a = 2.0 \, \text{m/s}^2 \) over a time interval \( t = 10 \, \text{s} \).

We use the following kinematic equation to calculate the distance traveled:

\[
d = v_0 t + \frac{1}{2} a t^2
\]

Substituting the known values into the equation:

\[
d = (5.0 \, \text{m/s})(10 \, \text{s}) + \frac{1}{2}(2.0 \, \text{m/s}^2)(10 \, \text{s})^2
\]

Calculating each term separately:

1. \( v_0 t = (5.0 \, \text{m/s})(10 \, \text{s}) = 50.0 \, \text{m} \)
2. \( \frac{1}{2} a t^2 = \frac{1}{2}(2.0 \, \text{m/s}^2)(100 \, \text{s}^2) = 1.0 \times 100 \, \text{m} = 100.0 \, \text{m} \)

Now, add the two distances together:

\[
d = 50.0 \, \text{m} + 100.0 \, \text{m} = 150.0 \, \text{m}
\]

Therefore, the total distance traveled by the car during the 10-second interval is \( \boxed{150.0 \, \text{m}} \).