Question

A car sits on an entrance ramp to a freeway, waiting for a break in the traffic. Then the driver accelerates with constant acceleration along the ramp and onto the freeway. The car starts from rest, moves in a straight line, and has a speed of 20 m/s (45 mi/h) when it reaches the end of the 120-m-long ramp. (a) What is the acceleration of the car? (b) How much time does it take the car to travel the length of the ramp? (c) The traffic on the freeway is moving at a constant speed of 20 m/s. What distance does the traffic travel while the car is moving the length of the ramp?

Short answer

(a) $a=\frac{5}{3}{\text{m/s}}^2$, (b) $t=12\text{s}$, (c) $d=240\text{m}$.

Step-by-step solution

Identify Known and Unknowns for Part (a)

The car starts from rest, which means its initial velocity $u=0$. The final velocity $v=20\text{m/s}$, and the distance $s=120\text{m}$. We need to find the acceleration $a$.

Use Equation of Motion for Acceleration

Using the equation $v^2=u^2+2as$, substitute the known values: $(20{)}^2=(0{)}^2+2a(120)$. Solving for $a$, we get $a=\frac{400}{240}=\frac{5}{3}{\text{m/s}}^2$.

Identify Known and Unknowns for Part (b)

Now, we need to find the time $t$ it takes to travel the ramp. We know $u=0$, $v=20\text{m/s}$, and $a=\frac{5}{3}{\text{m/s}}^2$.

Use Equation of Motion for Time

Using the equation $v=u+at$, substitute the known values: $20=0+\frac{5}{3}t$. Solving for $t$, we find $t=12\text{s}$.

Identify Known and Unknowns for Part (c)

The traffic on the freeway is moving at a constant speed of $20\text{m/s}$. We need to determine the distance traveled by the traffic in 12 seconds.

Calculate Distance Traveled by Traffic

Since distance $d=vt$ where $v=20\text{m/s}$ and $t=12\text{s}$, the distance $d=20\times 12=240\text{m}$.

Key concepts

constant acceleration

Constant acceleration means that the rate at which the speed or velocity of an object changes remains steady over time. In simpler terms, it means the object gains the same amount of speed every second. Imagine you are pressing down steadily on the pedal of a car. If you're accelerating at a constant rate, your speed increases smoothly, without any sudden spikes or drops.
Constant acceleration is an essential concept to understand because it simplifies the calculations of motion. When acceleration is constant:

• The behaviors of velocity and displacement become predictable.
• Motions can be described using basic equations.

Understanding constant acceleration helps in solving problems like the one in our exercise, where the car accelerates briskly up to freeway speed.

equations of motion

Equations of motion are vital tools in kinematics, the study of movement. These equations describe the mathematical relationship between displacement, velocity, acceleration, and time. For objects moving with constant acceleration, we have three primary equations:

• The first is: $v=u+at$
  It connects current velocity $v$, initial velocity $u$, acceleration $a$, and time $t$.
• The second is: $s=ut+\frac{1}{2}a{t}^2$
  This relates displacement $s$ with initial velocity, time, and acceleration.
• The third is: $v^2=u^2+2as$
  It shows the connection between velocities, acceleration, and displacement without involving time.

These equations allow you to solve for unknown motion values based on what is known. In our exercise, they help calculate the car’s acceleration and the time it takes to travel the ramp.

velocity and time

Velocity tells us how fast something is moving and in what direction. It's a vector quantity, meaning it includes both speed and direction. In the concept of motion under constant acceleration, velocity changes steadily.
When solving problems in kinematics, it's crucial to know either the initial or final velocity, along with the time, to find other unknowns. For example, using the equation $v=u+at$, if the car starts from rest ($u=0$) and accelerates to $20\text{m/s}$ over some time, we can determine both the acceleration and the time elapsed. Time is a fundamental factor:

• It's essential for calculating how long an object takes to get from one point to another.
• It also helps predict future motion states.

In our problem, understanding the timing helped determine how long the ramp journey took and calculated distances.

distance traveled

Distance traveled is a measure of the total length of the path taken by an object in motion. In straight-line motion with constant acceleration, calculating distance can be straightforward using motion equations.
The distance covered can be calculated using the expression:
$$s=ut+\frac{1}{2}a{t}^2$$
Here, $u$ is the initial velocity, $a$ is acceleration, and $t$ is the time duration of travel. In scenarios where speed is constant, the equation simplifies to:
$$d=vt$$
This is shown in our problem, where the steady-speed traffic on the freeway covers double the length of the ramp as the accelerating car, simply using its constant speed over time.

• Understanding how to compute distance provides insight into the motion and confirms other calculations.
• It's a vital concept when determining changes in an object's position over time.

By mastering the distance traveled, you gain a deeper grasp of how motion translates into real-world paths.