Question

A car and a truck start from rest at the same instant, with the car initially at some distance behind the truck. The truck has a constant acceleration of 2.10 m/s\(^2\), and the car has an acceleration of 3.40 m/s\(^2\). The car overtakes the truck after the truck has moved 60.0 m. (a) How much time does it take the car to overtake the truck? (b) How far was the car behind the truck initially? (c) What is the speed of each when they are abreast? (d) On a single graph, sketch the position of each vehicle as a function of time. Take \(x =\) 0 at the initial location of the truck.

Short answer

It takes 10.67 seconds for the car to overtake the truck, initially 19.2 meters behind. Their speeds are 22.4 m/s (truck) and 36.3 m/s (car) when abreast.

Step-by-step solution

Understanding the Problem

Let's break down what is given: the truck starts with an acceleration of 2.10 m/s² and moves 60.0 m, while the car starts from an unknown initial position with an acceleration of 3.40 m/s². We need to find the time it takes for the car to catch up with the truck, the initial distance between them, and their speeds when they are at the same position, and also sketch their positions over time.

Equation of Motion for the Truck

Use the equation of motion for the truck: \( x_t = \frac{1}{2} a_t t^2 \). Here, \( a_t = 2.10 \, \text{m/s}^2 \) and \( x_t = 60.0 \, \text{m} \). Thus, \( 60.0 = \frac{1}{2} \times 2.10 \times t^2 \). Simplify to find \( t \).

Solving for Time (Truck)

Solving \( 60.0 = 1.05 t^2 \) gives us \( t = \sqrt{\frac{60}{1.05}} \). Calculate this to find \( t \approx 10.67 \text{ seconds} \).

Equation of Motion for the Car

Use the same form of equation of motion for the car but consider an initial distance \( d \): \( x_c = d + \frac{1}{2} a_c t^2 \), where \( a_c = 3.40 \, \text{m/s}^2 \). When the car overtakes the truck, \( x_c = x_t = 60 \), and the time \( t \) is the same.

Solving for the Initial Distance

Set \( 60 = d + \frac{1}{2} \times 3.40 \times (10.67)^2 \). Solve for \( d \) to find \( d \approx 19.2 \, \text{m} \), meaning the car was initially 19.2 meters behind the truck.

Finding the Velocity of the Truck

Use the formula \( v_t = a_t t \) for the truck. Substitute \( a_t = 2.10 \) and \( t = 10.67 \) to find the velocity: \( v_t = 2.10 \times 10.67 = 22.4 \, \text{m/s} \).

Finding the Velocity of the Car

Use a similar formula \( v_c = a_c t \) for the car. Substitute \( a_c = 3.40 \) and \( t = 10.67 \) to find \( v_c = 3.40 \times 10.67 = 36.3 \, \text{m/s} \).

Sketching the Graphs

Plot the position vs. time for both the car and the truck. Start the truck at \((0, 0)\) and follow its quadratic trajectory \( x = 1.05 t^2 \). Start the car at \( (0, -19.2) \) and follow \( x = -19.2 + 1.7 t^2 \). Both should intersect at \( (10.67, 60) \).

Key concepts

Uniform Acceleration

Understanding uniform acceleration is key to solving kinematics problems like the one in our exercise. When an object accelerates uniformly, it means that the rate of change of its velocity is constant throughout its motion. This concept simplifies calculations since the acceleration value remains unchanged over time.

In the case of our truck and car problem, both the car and the truck start from rest, implying their initial velocity is zero. They accelerate at different constant rates of 2.10 m/s two for the truck and 3.40 m/s two for the car. This constant acceleration means we can reliably apply kinematic equations to predict their future positions and velocities based on time elapsed.

Uniform acceleration also allows us to use straightforward formulas, like the equations of motion and velocity-time graphs, to understand and solve the problem effectively.

Equations of Motion

The equations of motion are essential tools when dealing with problems involving uniform acceleration. These equations help relate the motion characteristics of an object, like displacement, time, initial velocity, and acceleration.

In our exercise, we apply one of the most fundamental equations of motion: \[ x = ut + \frac{1}{2} a t^2 \] Where:

• \( x \) is the displacement
• \( u \) is the initial velocity, which is zero for both the car and the truck
• \( a \) is the constant acceleration
• \( t \) is the time elapsed

For the truck, substituting into this equation allows us to find the time it takes to travel 60 meters. Similarly, for the car, this equation helps determine its position relative to the truck's position when both vehicles are side by side.

We also used the equation \( v = u + at \) to calculate the velocities of the car and truck afterward, which tells us how fast each was moving at the point of overtaking.

Velocity-time Graph

A velocity-time graph is a powerful visual tool that allows us to understand how the velocity of an object changes over time. It provides insights into the nature of an object's motion, especially when dealing with uniformly accelerated motion.

For objects with constant acceleration, like our car and truck, the velocity-time graph is a straight line. The slope of this line represents the acceleration. In this scenario, a steeper slope indicates a higher acceleration, which is why the car's line is steeper than the truck's.

The area under the velocity-time graph gives the displacement. For our problem, this information helps verify the calculations made using equations of motion. In essence, plotting these graphs for both vehicles, we can see how each progresses over time and pinpoint the moment they are side by side, reinforcing the algebraic solution with a visual check.