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Chapter 2: Problem 72

Two cars start 200 \(\mathrm{m}\) apart and drive toward each other at a steady 10 \(\mathrm{m} / \mathrm{s} .\) On the front of one of them, an energetic grasshopper jumps back and forth between the cars (he has strong legs!) with a constant horizontal velocity of 15 \(\mathrm{m} / \mathrm{s}\) relative to the ground. The insect jumps the instant he lands, so he spends no time resting on either car. What total distance does the grasshopper travel before the cars hit?

Short Answer

Expert verified
The grasshopper travels a total distance of 150 meters.

Step by step solution

01

Understanding the Scenario

Two cars are initially 200 meters apart and move towards each other, each at 10 m/s. There is a grasshopper that is traveling back and forth between the cars at a constant speed of 15 m/s, jumping instantly as it reaches a car.
02

Calculate Time Until Cars Collide

The combined speed of the two cars while they are moving toward each other is the sum of their speeds: 10 m/s + 10 m/s = 20 m/s. The original distance between the cars is 200 meters. To find the time it takes for the cars to collide, use the formula:\[ t = \frac{\text{distance}}{\text{combined speed}} = \frac{200}{20} = 10 \text{ seconds} \]
03

Determine Grasshopper's Total Distance Traveled

The grasshopper travels back and forth at a constant speed of 15 m/s. The time available for it to travel is the time until the cars collide, which we've found to be 10 seconds.The total distance the grasshopper travels is given by:\[ ext{Distance} = ext{speed} \times ext{time} = 15 \times 10 = 150 \text{ meters} \]

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Key Concepts

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

Understanding Relative Velocity
Relative velocity is a key concept in kinematics, particularly when dealing with multiple moving objects. In this scenario, imagine two cars moving toward each other with a grasshopper jumping between them. The cars each have a velocity of 10 meters per second (m/s), and this velocity is measured relative to the ground. Since both cars move towards each other, their relative velocity when considered together is the sum of their individual speeds.
  • Car 1 velocity: 10 m/s towards Car 2
  • Car 2 velocity: 10 m/s towards Car 1
  • Relative velocity of the cars: 10 m/s + 10 m/s = 20 m/s
The grasshopper adds another layer. It moves with a constant velocity of 15 m/s relative to the ground. The crucial point here is that the grasshopper鈥檚 relative speed compared to each car varies as it jumps between them. However, for calculating total travel, we only care about its speed relative to the ground, consistently given as 15 m/s. This understanding of relative velocity simplifies the situation, allowing us to focus on linear motion without complications.
Calculating Collision Time
To determine when two moving objects will collide, the time calculation is based on their relative velocity and the initial distance between them. In our example, the two cars start 200 meters apart and move towards each other at a combined speed of 20 m/s. The formula to find the collision time is \[ t = \frac{\text{distance}}{\text{combined speed}}\]Inserting our values:\[t = \frac{200\, \text{m}}{20\, \text{m/s}} = 10\, \text{seconds}\]This result tells us that the two cars will meet after 10 seconds. Understanding this time frame is vital as it tells us how long any other events can occur within that period, like the grasshopper鈥檚 leaps. By knowing how long the cars take to collide, we can figure out how far the grasshopper travels鈥攚hich, in essence, is sustained during this precise time window.
Concept of Constant Speed Motion
Constant speed motion is when an object travels at a fixed speed over time. Here, the grasshopper moves at a constant speed of 15 m/s. It continuously jumps between the cars, never slowing down or stopping until the cars collide. This consistent movement without rest means that the total distance it covers depends directly on the collision time previously calculated.The formula for determining distance in constant speed motion scenarios is:\[\text{Distance} = \text{Speed} \times \text{Time}\]Given the grasshopper鈥檚 speed and the time until collision is 10 seconds, the computation becomes:\[\text{Distance} = 15\, \text{m/s} \times 10\, \text{s} = 150\, \text{m}\]This calculation indicates the grasshopper travels a total of 150 meters. The constant speed motion simplifies our task as it eliminates concerns about acceleration or deceleration, making the prediction of travel straightforward once the critical parameters, like speed and time, are known.

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Most popular questions from this chapter

Large cockroaches can run as fast as 1.50 \(\mathrm{m} / \mathrm{s}\) in short bursts. Suppose you turn on the light in a cheap motel and see one scurrying directly away from you at a constant 1.50 \(\mathrm{m} / \mathrm{s} .\) If you start 0.90 m behind the cockroach with an initial speed of 0.80 \(\mathrm{m} / \mathrm{s}\) toward it, what minimum constant acceleration would you need to catch up with it when it has traveled \(1.20 \mathrm{m},\) just short of safety under a counter?

A rocket starts from rest and moves upward from the surface of the earth. For the first 10.0 s of its motion, the vertical acceleration of the rocket is given by \(a_{y}=\left(2.80 \mathrm{m} / \mathrm{s}^{3}\right) t,\) where acceleration of the rocket is given by \(a_{y}=\left(2.80 \mathrm{m} / \mathrm{s}^{3}\right) t,\) where the \(+y\) -direction is upward. (a) What is the height of the rocket above the surface of the earth at \(t=10.0 \mathrm{s} ?\) (b) What is the speed of the rocket when it is 325 \(\mathrm{m}\) above the surface of the earth?

Two cars, \(A\) and \(B,\) travel in a straight line. The dis tance of \(A\) from the starting point is given as a function of time by \(x_{A}(t)=\alpha t+\beta t^{2},\) with \(\alpha=2.60 \mathrm{m} / \mathrm{s}\) and \(\beta=1.20 \mathrm{m} / \mathrm{s}^{2} .\) The distance of \(B\) from the starting point is \(x_{B}(t)=\gamma t^{2}-\delta t^{3},\) with \(\gamma=2.80 \mathrm{m} / \mathrm{s}^{2}\) and \(\delta=0.20 \mathrm{m} / \mathrm{s}^{3} .\) (a) Which car is ahead just after they leave the starting point? (b) At what time(s) are the cars at the same point? (c) At what time(s) is the distance from \(A\) to \(B\) neither increasing nor decreasing? (d) At what time(s) do \(A\) and \(B\) have the same acceleration?

Determined to test the law of gravity for himself, a student walks off a skyscraper 180 \(\mathrm{m}\) high, stopwatch in hand, and starts his free fall (zero initial velocity). Five seconds later, Superman arrives at the scene and dives off the roof to save the student. Superman leaves the roof with an initial speed \(v_{0}\) that he produces by pushing himself downward from the edge of the roof with his legs of steel. He then falls with the same acceleration as any freely falling body. (a) What must the value of \(v_{0}\) be so that Superman catches the student just before they reach the ground? (b) On the same graph, sketch the positions of the student and of Superman as functions of time. Take Superman's initial speed to have the value calculated in part (a). (c) If the height of the skyscraper is less than some minimum value, even Superman can't reach the student before he hits the ground. What is this minimum height?

Prevention of Hip Fractures. Falls resulting in hip fractures are a major cause of injury and even death to the elderly. Typically, the hip's speed at impact is about 2.0 \(\mathrm{m} / \mathrm{s} .\) If this can be reduced to 1.3 \(\mathrm{m} / \mathrm{s}\) or less, the hip will usually not fracture. One way to do this is by wearing elastic hip pads. (a) If a typical pad is 5.0 \(\mathrm{cm}\) thick and compresses by 2.0 \(\mathrm{cm}\) during the impact of a fall, what constant acceleration \(\operatorname{m} / \mathrm{s}^{2}\) and in \(g^{\prime} \mathrm{s}\) ) does the hip undergo to reduce its speed from 2.0 \(\mathrm{m} / \mathrm{s}\) to 1.3 \(\mathrm{m} / \mathrm{s} ?\) (b) The acceleration you found in part (a) may seem rather large, but to fully assess its effects on the hip, calculate how long it lasts.
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