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Chapter 2: Problem 4

A truck's speed increases uniformly from \(15 \mathrm{~km} / \mathrm{h}\) to \(60 \mathrm{~km} / \mathrm{h}\) in \(20 \mathrm{~s}\). Determine \((a)\) the average speed, \((b)\) the acceleration, and \((c)\) the distance traveled, all in units of meters and seconds. For the 20-s trip under discussion, taking \(+x\) to be in the direction of motion, $$\begin{array}{l} v_{i x}=\left(15 \frac{\mathrm{km}}{\mathrm{K}}\right)\left(1000 \frac{\mathrm{m}}{\mathrm{k} \mathrm{m}}\right)\left(\frac{1}{3600} \frac{\mathrm{K}}{\mathrm{s}}\right)=4.17 \mathrm{~m} / \mathrm{s} \\ v_{f x}=60 \mathrm{~km} / \mathrm{h}=16.7 \mathrm{~m} / \mathrm{s} \end{array}$$ (a) \(v_{a v}=\frac{1}{2}\left(v_{i x}+v_{f x}\right)=\frac{1}{2}(4.17+16.7) \mathrm{m} / \mathrm{s}=10 \mathrm{~m} / \mathrm{s}\) (b) \(\quad a=\frac{v_{f x}-v_{i x}}{t}=\frac{(16.7-4.17) \mathrm{m} / \mathrm{s}}{20 \mathrm{~s}}=0.63 \mathrm{~m} / \mathrm{s}^{2}\) (c) \(\quad x=v_{a v} t=(10.4 \mathrm{~m} / \mathrm{s})(20 \mathrm{~s})=208 \mathrm{~m}=0.21 \mathrm{~km}\)

Short Answer

Expert verified
The average speed is 10.4 m/s, acceleration is 0.63 m/s虏, and distance traveled is 208 m.

Step by step solution

01

Convert Initial and Final Speeds

Initially, the truck's speed is given in km/h, so we must convert everything to m/s. We use the conversion: \(1 \text{ km/h} = \frac{1}{3.6} \text{ m/s}\). Therefore, \(v_{ix} = 15 \times \frac{1000}{3600} = 4.17 \text{ m/s}\) and \(v_{fx} = 60 \times \frac{1000}{3600} = 16.7 \text{ m/s}\).
02

Calculate Average Speed

The average speed during uniform acceleration is the average of the initial and final speeds. We use the formula: \(v_{av} = \frac{v_{ix} + v_{fx}}{2}\). Substituting the values, \(v_{av} = \frac{4.17 + 16.7}{2} = 10.4 \text{ m/s}\).
03

Determine Acceleration

Using the formula for acceleration, \(a = \frac{v_{fx} - v_{ix}}{t}\), we find \(a = \frac{16.7 - 4.17}{20} = 0.63 \text{ m/s}^2\).
04

Find Distance Traveled

We calculate the distance using the average speed and time: \(x = v_{av} \times t\). Substituting the values, \(x = 10.4 \times 20 = 208 \text{ m}\).

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Key Concepts

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

Average Speed
When we talk about average speed, we're referring to the total distance traveled divided by the total time of travel. This measurement is especially helpful when the speed is changing, such as with uniform acceleration. For our problem, average speed is the mean of the truck's initial and final velocities over the period in question. Since the speed increases uniformly, we can easily calculate it using the formula:\[v_{av} = \frac{v_{i} + v_{f}}{2}\]Here, the truck's initial speed (\(v_{i}\)) is 4.17 m/s, and its final speed (\(v_{f}\)) is 16.7 m/s. Plugging these values into the formula gives us:\[v_{av} = \frac{4.17 + 16.7}{2} = 10.4 \text{ m/s}\]This value of 10.4 m/s represents the truck's average speed during the 20-second increase in velocity. It simplifies understanding how fast the truck was traveling throughout the interval, despite its changing speed.
Acceleration Calculation
Acceleration indicates how quickly an object changes its speed. In scenarios dealing with uniform acceleration, like in this problem, the object speeds up at a constant rate. We calculate acceleration using the formula:\[a = \frac{v_{f} - v_{i}}{t}\]Here, \(v_{f}\) is the final velocity (16.7 m/s), \(v_{i}\) is the initial velocity (4.17 m/s), and \(t\) is the time interval (20 seconds). Using the values provided:\[a = \frac{16.7 - 4.17}{20} = 0.63 \text{ m/s}^{2}\]This result, 0.63 m/s虏, tells us the rate at which the truck's speed increased every second over the 20-second period. Understanding this helps in evaluating how the object's velocity changes with time under constant acceleration conditions.
Distance Traveled
Distance traveled is a straightforward concept, but it can get a bit more complex when speeds change. With uniform acceleration, we can determine the distance by using the average speed and the total time. Our distance formula is:\[x = v_{av} \times t\]Where the average speed \(v_{av}\) is 10.4 m/s, and \(t\) is the time duration of 20 seconds. Inputting these values, we find:\[x = 10.4 \times 20 = 208 \text{ meters}\]The truck covers a distance of 208 meters during the 20-second interval. Distance calculations like this are valuable as they give a clear picture of how far an object moves over time, even when its speed is changing uniformly.

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Most popular questions from this chapter

A ball is dropped from rest at a height of \(50 \mathrm{~m}\) above the ground. (a) What is its speed just before it hits the ground? \((b)\) How long does it take to reach the ground? If we can ignore air friction, the ball is uniformly accelerated until it reaches the ground. Its acceleration is downward and is \(9.81 \mathrm{~m} / \mathrm{s}^{2}\). Taking down as positive, we have for the trip: $$y=50.0 \mathrm{~m} \quad a=9.81 \mathrm{~m} / \mathrm{s}^{2} \quad v_{i}=0$$ (a) \(v_{f y}^{2}=v_{i y}^{2}+2 a y=0+2\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(50.0 \mathrm{~m})=981 \mathrm{~m}^{2} / \mathrm{s}^{2}\) and so \(v_{f}=31.3 \mathrm{~m} / \mathrm{s}\). (b) From \(a\left(v_{f y}-v_{i y}\right) / t\), $$t=\frac{v_{f y}-v_{i y}}{a}=\frac{(31.3-0) \mathrm{m} / \mathrm{s}}{9.81 \mathrm{~m} / \mathrm{s}^{2}}=3.19 \mathrm{~s}$$ (We could just as well have taken \(u p\) as positive. How would the calculation have been changed?)

A ballast bag is dropped from a balloon that is \(300 \mathrm{~m}\) above the ground and rising at \(13 \mathrm{~m} / \mathrm{s}\). For the bag, find \((a)\) the maximum height reached, \((b)\) its position and velocity \(5.0 \mathrm{~s}\) after it is released, and \((c)\) the time at which it hits the ground. The initial velocity of the bag when released is the same as that of the balloon, \(13 \mathrm{~m} / \mathrm{s}\) upward. Choose \(u p\) as positive and take \(y=0\) at the point of release. (a) At the highest point, \(v_{f}=0\). From \(v_{f y}^{2}=v_{i y}^{2}+2 a y\), $$0=(13 \mathrm{~m} / \mathrm{s})^{2}+2\left(-9.81 \mathrm{~m} / \mathrm{s}^{2}\right) y \quad \text { or } \quad y=8.6 \mathrm{~m}$$ The maximum height is \(300+8.6=308.6 \mathrm{~m}\) or \(0.31 \mathrm{~km}\). (b) Take the end point to be its position at \(t=5.0 \mathrm{~s}\). Then, from \(y=v_{i y} t+\frac{1}{2} a t^{2}\), $$y=(13 \mathrm{~m} / \mathrm{s})(5.0 \mathrm{~s})+\frac{1}{2}\left(-9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(5.0 \mathrm{~s})^{2}=-57.6 \mathrm{~m} \text { or }-58 \mathrm{~m}$$ So its height is \(300-58=242 \mathrm{~m}\). Also, from \(v_{f y}=v_{i y}+a t\), $$v_{f y}=13 \mathrm{~m} / \mathrm{s}+\left(-9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(5.0 \mathrm{~s})=-36 \mathrm{~m} / \mathrm{s}$$ It is on its way down with a velocity of \(36 \mathrm{~m} / \mathrm{s}-\) DOWNWARD. (c) Just as it hits the ground, the bag's displacement is \(-300 \mathrm{~m}\). Then \(y=v_{i y} t+\frac{1}{2} a t^{2} \quad\) becomes \(\quad-300 \mathrm{~m}=(13 \mathrm{~m} / \mathrm{s}) t+\frac{1}{2}\left(-9.81 \mathrm{~m} / \mathrm{s}^{2}\right) t^{2}\) or \(4.905 t^{2}-13 t-300=0\). The quadratic formula gives \(t=9.3 \mathrm{~s}\) and \(-6.6 \mathrm{~s}\). Only the positive time has physical meaning, so the required answer is \(9.3 \mathrm{~s}\). as before.

A body is projected downward at an angle of \(30^{\circ}\) with the horizontal from the top of a building \(170 \mathrm{~m}\) high. Its initial speed is \(40 \mathrm{~m} / \mathrm{s}\). (a) How long will it take before striking the ground? \((b)\) How far from the foot of the building will it strike? ( \(c\) ) At what angle with the horizontal will it strike?

A baseball is thrown straight upward on the Moon with an initial speed of \(35 \mathrm{~m} / \mathrm{s}\). Compute \((a)\) the maximum height reached by the ball, \((b)\) the time taken to reach that height, \((c)\) its velocity \(30 \mathrm{~s}\) after it is thrown, and \((d)\) when the ball's height is \(100 \mathrm{~m}\). Take \(u p\) as positive. At the highest point, the ball's velocity is zero. (a) From \(v_{f y}^{2}=v_{i y}^{2}+2 a y\), since \(g=1.60 \mathrm{~m} / \mathrm{s}^{2}\) on the Moon, $$0=(35 \mathrm{~m} / \mathrm{s})^{2}+2\left(-1.60 \mathrm{~m} / \mathrm{s}^{2}\right) y \quad \text { or } \quad y=0.38 \mathrm{~km}$$ (b) From \(v_{f y}=v_{i y}+a t\) $$0=35 \mathrm{~m} / \mathrm{s}+\left(-1.60 \mathrm{~m} / \mathrm{s}^{2}\right) t \quad \text { or } \quad t=22 \mathrm{~s}$$ (c) From \(v_{f y}=v_{i y}+a t\) $$v_{f y}=35 \mathrm{~m} / \mathrm{s}+\left(-1.60 \mathrm{~m} / \mathrm{s}^{2}\right)(30 \mathrm{~s}) \quad \text { or } \quad v_{f y}=-13 \mathrm{~m} / \mathrm{s}$$ Because \(v_{f}\) is negative and we are taking \(u p\) as positive, the velocity is directed downward. The ball is on its way down at \(t=30 \mathrm{~s}\). (d) From \(y=v_{i y} t+\frac{1}{2} a t^{2}\) we have $$100 \mathrm{~m}=(34 \mathrm{~m} / \mathrm{s}) t+\frac{1}{2}\left(-1.60 \mathrm{~m} / \mathrm{s}^{2}\right) t^{2} \quad \text { or } \quad 0.80 t^{2}-35 t+100=0$$ By use of the quadratic formula, or $$\begin{aligned} x &=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \\ t &=\frac{35 \pm \sqrt{35^{2}-4(0.80) 100}}{2(0.80)}=\frac{35 \pm 30.08}{1.60} \end{aligned}$$ we find \(t=3.1 \mathrm{~s}\) and \(41 \mathrm{~s}\). At \(t=3.1 \mathrm{~s}\) the ball is at \(100 \mathrm{~m}\) and ascending; at \(t=41 \mathrm{~s}\) it is at the same height but descending.

A World Series batter hits a home run ball with a velocity of \(40 \mathrm{~m} / \mathrm{s}\) at an angle of \(26^{\circ}\) above the horizontal. A fielder who can reach \(3.0 \mathrm{~m}\) above the ground is backed up against the bleacher wall, which is \(110 \mathrm{~m}\) from home plate. The ball was \(120 \mathrm{~cm}\) above the ground when hit. How high above the fielder's glove does the ball pass?
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