Problem 33 A car moving with constant accel... [FREE SOLUTION] (2024)

Short Answer

equations of motion

The equations of motion describe how a physical object's position, speed, and acceleration change over time. They are especially useful for problems involving constant acceleration. The three classic equations are:
1. \( v = u + at \): This calculates velocity (v) at time t given the initial velocity (u) and acceleration (a).
2. \( s= ut + \frac{1}{2}at^2 \): This calculates distance (s) traveled.
3. \( v^2 = u^2 + 2as \): This relates the final velocity to initial velocity, acceleration, and distance traveled.
In our exercise, these equations help determine initial speed, acceleration, and the location where the car was initially at rest.

initial velocity

To find the car's initial velocity (\( V_1 \)), we start from a given distance and final velocity after a specific time. The car covers 60.0 meters in 6.00 seconds, reaching 15.0 m/s at the end. Using the equation:
\[ s = \frac{(V_1 + V_2)}{2} \times t \]
where s = 60.0 m, V_2 = 15.0 m/s, and t = 6.00 s. Rearranging to solve for V_1:
\[ V_1 = \frac{2s}{t} - V_2 \rightarrow \frac{2 \times 60.0}{6.00} - 15.0 = 5.0 \text{ m/s} \]
Thus, the initial speed of the car was 5.0 m/s.

acceleration calculation

Acceleration (\( a \)) indicates how quickly the car's speed changes. Knowing the initial and final velocities and the time it took, we use:
\[ V_2 = V_1 + at \]
Rearranging to solve for acceleration:
\[ a = \frac{(V_2 - V_1)}{t} \rightarrow \frac{(15.0 - 5.0)}{6.00} = 1.67 \text{ m/s}^2 \]
So, the car's acceleration is 1.67 m/s^2.

distance calculation

We need to find what distance the car covered from a point of rest to reach its initial speed of 5.0 m/s. Using the equation:
\[ V_2^2 = V_1^2 + 2as \]
Here, V_2 is the initial speed in the previous calculation (5.0 m/s), and V_1 is 0 since the car was at rest. Rearranging to solve for s:
\[ s = \frac{V_1^2}{2a} \rightarrow \frac{5.0^2}{2 \times 1.67} = 7.5 \text{ m} \]
The car was at rest 7.5 meters before the calculated initial point.

graphical representation of motion

Graphs provide a clear visual representation of the car's motion. For position vs. time (x vs. t), use:
\[ x = V_1 t + \frac{1}{2}at^2 \]
which forms a parabola. For velocity vs. time (v vs. t), use:
\[ v = V_1 + at \]
which creates a linear graph. Plot these to observe acceleration's effects: