Question

At the instant the traffic light turns green, a car that has been waiting at an intersection starts ahead with a constant acceleration of 2.80 m/s\(^2\). At the same instant a truck, traveling with a constant speed of 20.0 m/s, overtakes and passes the car. (a) How far beyond its starting point does the car overtake the truck? (b) How fast is the car traveling when it overtakes the truck? (c) Sketch an \(x-t\) graph of the motion of both vehicles. Take \(x=\) 0 at the intersection. (d) Sketch a \(v_x-t\) graph of the motion of both vehicles.

Short answer

(a) 285.7 meters. (b) 40 m/s. (c) & (d) Refer to the graph descriptions.

Step-by-step solution

Establish Car and Truck Equations

For the car, starting from rest with a constant acceleration, the motion equation is given by \( x_c = \frac{1}{2} a t^2 \), where \( a = 2.8 \text{ m/s}^2 \). Hence, \( x_c = 1.4 t^2 \).For the truck moving with a constant speed, its motion equation is \( x_t = v_t t \), where \( v_t = 20.0 \text{ m/s} \). Therefore, \( x_t = 20t \).

Set Equations Equal to Find Time

To find when the car overtakes the truck, set \( x_c = x_t \). This gives the equation \( 1.4t^2 = 20t \). Solve this by first dividing both sides by \( t \) (since \( t eq 0 \)) to obtain \( 1.4t = 20 \). Thus, \( t = \frac{20}{1.4} \approx 14.29 \text{ seconds} \).

Calculate Distance Overtaken

Substitute \( t = 14.29 \) back into any of the position equations, let's use \( x_c = 1.4 t^2 \):\[ x_c = 1.4 \times (14.29)^2 \approx 285.7 \text{ meters} \].Thus, the car overtakes the truck 285.7 meters beyond its starting point.

Calculate Speed of the Car

The car's velocity at the time it overtakes the truck is given by \( v_c = a t \).Using \( a = 2.80 \text{ m/s}^2 \) and \( t = 14.29 \text{ seconds} \):\[ v_c = 2.80 \times 14.29 \approx 40 \text{ m/s} \].

Draw the Distance-Time Graph

On the distance-time graph, plot the truck's path as a straight line and the car's path as a parabola originating from the origin. The intersection of these two curves indicates the position and time when the car overtakes the truck.

Draw the Velocity-Time Graph

On the velocity-time graph, represent the car's motion as a straight line with a positive slope (starting from zero) due to its constant acceleration. The truck's motion is represented as a horizontal line since its velocity is constant.

Key concepts

Constant Acceleration

In kinematics, one of the interesting scenarios is analyzing motion with constant acceleration. This means that the velocity of an object changes at a constant rate over time. For example, when a car accelerates from rest, it gradually increases its speed. If the car in our case accelerates at 2.80 m/s\(^2\), it gains speed consistently. The equation to describe the position of such an object over time is given by:
\[x = x_0 + v_0 t + \frac{1}{2} a t^2\]
For our scenario, the car starts from rest, simplifying it to \( x = \frac{1}{2} a t^2 \), as the initial velocity \( v_0 = 0 \). Constant acceleration equations are powerful tools for predicting motion in physics. They provide insights into how fast an object moves and the distance it covers over time.

Velocity-Time Graph

A velocity-time graph is a useful tool in understanding motion. It depicts how an object's speed changes over time. For our example, the velocity-time graph of the car, which has constant acceleration, starts at the origin and ascends in a straight line. This straight line signifies a steady increase in velocity.

• The slope of this line represents acceleration—steeper slopes indicate higher acceleration.
• For the car with 2.80 m/s\(^2\) acceleration, its graph is a straight line with a positive slope.

Conversely, the truck moves at a constant speed of 20.0 m/s. Its velocity-time graph is a horizontal line, meaning its speed doesn’t change over time. The differences in these representations make it easier to understand how different vehicles behave over the same period.

Motion Equations

Motion equations are essential in solving kinematics problems. They link displacement, velocity, acceleration, and time. One of the fundamental equations when dealing with constant acceleration is:
\[v = v_0 + at\]
It helps in finding the velocity of an object at any given time. In our context, this equation tells us the car’s velocity when it eventually overtakes the truck.

• By plugging in the values, we find that the car's velocity when it catches up to the truck is approximately 40 m/s.
• This is derived using its constant acceleration and the time it has been accelerating.

These equations form the bedrock of analyzing motion scenarios, allowing us to calculate various unknowns given some known variables.

Distance-Time Graph

Distance-time graphs are visual representations of an object's position over time. They reveal how far an object travels as time progresses. For the car accelerating away from rest, the graph is a curve opening upward, a parabolic shape due to the quadratic nature of the equation \( x = \frac{1}{2}at^2 \).

• The curve starts at the origin and its steepness increases, representing the car's increasing speed over time.
• At any given point, the slope of the curve corresponds to the car's velocity at that time.

The truck, however, travels at a constant speed, so its distance-time graph is a straight, diagonal line starting from the intersection. The point where the car’s curve intersects the truck’s line is the exact location and time where the car overtakes the truck. This intersection is key in understanding relative motion between the two vehicles.