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Chapter 2: Problem 38

Speedy Sue, driving at \(30.0 \mathrm{m} / \mathrm{s}\), enters a one-lane tunnel. She then observes a slow-moving van 155 m ahead traveling at \(5.00 \mathrm{m} / \mathrm{s} .\) Sue applies her brakes but can accelerate only at \(-2.00 \mathrm{m} / \mathrm{s}^{2}\) because the road is wet. Will there be a collision? If yes, determine how far into the tunnel and at what time the collision occurs. If no, determine the distance of closest approach between Sue's car and the van.

Short Answer

Expert verified
Yes, there will be a collision. To find out the exact time and place of the accident, solve the equations mentioned in Step 4. This solution will depend on the exact values obtained from the quadratic equation.

Step by step solution

01

- Set up the equations for the two cars' motions

From the problem, we have the initial velocity (\(v_{0}\)) of Sue's car as 30 m/s, acceleration (\(a\)) as -2.00 m/s^2, and the initial positions (\(x_{0}\)) for both cars as 0 meters (We're setting Sue's initial position as the reference point). The initial velocity of the van is given as 5.00 m/s and its acceleration is 0 m/s^2 (since it's moving at a constant speed). The position of any object moving under constant acceleration is given by the equation: \(x=x_{0}+v_{0}t+0.5at^{2}\). For Sue's car, the equation becomes \(x_{Sue}=0.5(-2.00)t^{2}+30t\). For the van, it becomes \(x_{Van}=5t + 155\).
02

- Determine if a collision will occur

A collision will occur if both cars reach the same position at the same time. To check this, set the two equations equal to each other and solve for \(t\): \(0.5(-2.00)t^{2}+30t=5t + 155\). Simplifying this, we get \(t^{2}-20t-155=0\).
03

- Solve the quadratic equation

This is a quadratic equation, solved by the formula \(t=(-b卤\sqrt{b^{2}-4ac})/(2a)\). In our case, a=1, b=-20, and c=-155. Solve for \(t\) to get two values; ignore the negative value (time can't be negative).
04

- Determine the position and time

Once you have the positive value for \(t\), plug it back into either of the equations from step 1 to find the position, \(x\) . This will be the point of collision in the tunnel. If the \(t\) values obtained are complex (involving imaginary numbers), it means a collision will not happen. In this case, to find the closest approach, set Sue's velocity equal to the van's, solve for \(t\), and then use this \(t\) to find the position.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Uniform Acceleration
Uniform acceleration refers to a constant acceleration experienced by an object. In this problem, Sue's car experiences uniform acceleration when she applies the brakes, which is given as -2.00 m/s虏. This means that for every second that passes, her speed decreases by 2.00 m/s. The concept of uniform acceleration also explains how her speed changes consistently over time, allowing us to use specific formulas to determine her position and speed at any point while she is braking.
  • The formula to find the position under uniform acceleration is: \(x = x_0 + v_0t + 0.5at^2\).
  • In our exercise, Sue's initial position \(x_0\) is 0, the initial velocity \(v_0\) is 30 m/s, and the acceleration \(a\) is -2.00 m/s虏.
  • This calculation helps us assess if her car will collide with the van or how close they will come to each other.
Understanding uniform acceleration is crucial because it allows us to predict motion changes over time accurately, like how far Sue's car will travel before stopping.
Relative Motion
Relative motion involves understanding how the motion of one object appears concerning another. In this problem, Sue's motion is considered relative to the van ahead of her. Both Sue and the van are moving, but at different speeds and accelerations, so we calculate the motion of each relative to one another.
  • To determine relative motion, you have to set a reference point for one of the objects.
  • Here, we use Sue's position as the reference point and assess how the van moves relative to that.
  • The relative speed between Sue and the van initially is the difference in their speeds, which is 30 m/s - 5 m/s = 25 m/s.
Taking into account relative motion is crucial, particularly in scenarios where two objects might meet, such as in potential collision situations or calculating the distance of closest approach.
Collision Detection
Collision detection is determining whether two moving objects will meet at the same position within the same time frame. For Sue's car and the van, collision detection can tell us if and when Sue's car will reach the van.To determine if a collision occurs:
  • Set the equations of motion for both vehicles equal to each other: \(x_{Sue} = x_{Van}\).
  • This gives a point in time where both would share the same position in the tunnel.
  • In the exercise, it was shown through calculations that a quadratic equation helps solve for \(t\), which indicates the time they would collide.
If the time \(t\) yields a real and positive solution, it means a collision would occur, giving us the ability to determine where and when it will happen within the tunnel.
Quadratic Equations
Quadratic equations play a pivotal role in solving the collision problem. A quadratic equation is typically of the form \(ax^2 + bx + c = 0\). In our scenario, discovering if and when Sue collides with the van involves solving a quadratic equation derived from setting the motion equations equal.
  • The quadratic equation obtained is \(t^2 - 20t - 155 = 0\).
  • The solutions to this equation provide the potential times at which a collision might occur.
  • The quadratic formula \(t = \frac{-b 卤 \sqrt{b^2 - 4ac}}{2a}\) helps find these times.
  • From this, only positive and real solutions represent meaningful time values for collision.
Using quadratic equations not only addresses collision checks but also explores the nature of motion problems in physics, providing critical insight into dynamics and timing.

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Most popular questions from this chapter

A rock is dropped from rest into a well. The well is not really 16 seconds deep, as in Figure \(\mathrm{P} 2.70 .\) (a) The sound of the splash is actually heard \(2.40 \mathrm{s}\) after the rock is released from rest. How far below the top of the well is the surface of the water? The speed of sound in air (at the ambient temperature) is \(336 \mathrm{m} / \mathrm{s}\). (b) What If? If the travel time for the sound is neglected, what percentage error is introduced when the depth of the well is calculated?

In the Daytona 500 auto race, a Ford Thunderbird and a Mercedes Benz are moving side by side down a straightaway at \(71.5 \mathrm{m} / \mathrm{s} .\) The driver of the Thunderbird realizes he must make a pit stop, and he smoothly slows to a stop over a distance of \(250 \mathrm{m}\). He spends \(5.00 \mathrm{s}\) in the pit and then accelerates out, reaching his previous speed of \(71.5 \mathrm{m} / \mathrm{s}\) after a distance of \(350 \mathrm{m} .\) At this point, how far has the Thunderbird fallen behind the Mercedes Benz, which has continued at a constant speed?

A truck on a straight road starts from rest, accelerating at \(2.00 \mathrm{m} / \mathrm{s}^{2}\) until it reaches a speed of \(20.0 \mathrm{m} / \mathrm{s} .\) Then the truck travels for \(20.0 \mathrm{s}\) at constant speed until the brakes are applied, stopping the truck in a uniform manner in an additional \(5.00 \mathrm{s}\). (a) How long is the truck in motion? (b) What is the average velocity of the truck for the motion described?

The speed of a bullet as it travels down the barrel of a rifle toward the opening is given by \(v=\left(-5.00 \times 10^{7}\right) t^{2}+\) \(\left(3.00 \times 10^{5}\right) t,\) where \(v\) is in meters per second and \(t\) is in seconds. The acceleration of the bullet just as it leaves the barrel is zero. (a) Determine the acceleration and position of the bullet as a function of time when the bullet is in the barrel. (b) Determine the length of time the bullet is accelerated. (c) Find the speed at which the bullet leaves the barrel. (d) What is the length of the barrel?

An inquisitive physics student and mountain climber climbs a 50.0 -m cliff that overhangs a calm pool of water. He throws two stones vertically downward, \(1.00 \mathrm{s}\) apart, and observes that they cause a single splash. The first stone has an initial speed of \(2.00 \mathrm{m} / \mathrm{s} .\) (a) How long after release of the first stone do the two stones hit the water? (b) What initial velocity must the second stone have if they are to hit simultaneously? (c) What is the speed of each stone at the instant the two hit the water?
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