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Chapter 2: Problem 81

A rocket is launched straight up with constant acceleration. Four seconds after liftoff, a bolt falls off the side of the rocket. The bolt hits the ground 6.0 s later. What was the rocket's acceleration?

Short Answer

Expert verified
The rocket's acceleration was 9.8 m/s虏.

Step by step solution

01

Find the total time of flight for the bolt

The total time for which the bolt was in motion can be found by adding the time after which it fell (4 seconds) and the time it took to hit the ground (6 seconds) which totals to \( t = 4s + 6s = 10s \)
02

Use the equation of motion to find the velocity

We use the formula for the final velocity \( v = u + at \), where \( v \) is the velocity when the bolt hits the ground, \( u \) is the initial velocity (0, since the bolt began its motion at rest), \( a \) is the acceleration (-9.8 m/s虏, since it's moving against gravity), and \( t \) is the total time in flight. Substituting the values, we get \( v = 0 - 9.8m/s虏 \times 10s \). This gives the result \( v = -98 m/s \). This negative value indicates that the bolt was moving in the downward direction.
03

Find the velocity of the rocket when the bolt fell off

Since we know the total time and velocity, we can find out the velocity of the rocket when the bolt fell off. The bolt was in the air for 4 seconds before it started to fall, hence the velocity at this point can be calculated using the same motion equation, \( v = u + at \), but this time \( t = 4s \). Substituting the values, we get \( v = 0 - 9.8 m/s虏 \times 4s = -39.2 m/s \)
04

Calculate the rocket's acceleration

As per Newton's first law of motion, the bolt was moving with the same velocity as the rocket was, when it fell off. Therefore, the rocket was moving with an upward velocity of 39.2 m/s when the bolt fell. This velocity was achieved in 4 seconds. Using the formula for acceleration \( a = \frac {v - u} {t} \), where \( v \) is the final velocity, \( u \) is the initial velocity (0, as the rocket started from rest), and \( t \) is the time taken. Substituting values, we get \( a = \frac {39.2 m/s - 0} {4s} = 9.8 m/s虏 \). Hence the rocket's acceleration is 9.8 m/s虏.

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Most popular questions from this chapter

A Porsche challenges a Honda to a \(400 \mathrm{m}\) race. Because the Porsche's acceleration of \(3.5 \mathrm{m} / \mathrm{s}^{2}\) is larger than the Honda's \(3.0 \mathrm{m} / \mathrm{s}^{2},\) the Honda gets a \(100-\mathrm{m}\) head start-it is only \(300 \mathrm{m}\) from the finish line. Assume, somewhat unrealistically, that both cars can maintain these accelerations the entire distance. Who wins, and by how much time?

A cross-country skier is skiing along at a zippy \(8.0 \mathrm{m} / \mathrm{s}\). She stops pushing and simply glides along, slowing to a reduced speed of \(6.0 \mathrm{m} / \mathrm{s}\) after gliding for \(5.0 \mathrm{m}\). What is the magnitude of her acceleration as she slows?

A student at the top of a building of height \(h\) throws ball \(A\) straight upward with speed \(v_{0}\) and throws ball B straight downward with the same initial speed. a. Compare the balls" accelerations, both direction and magnitude, immediately after they leave her hand. Is one acceleration larger than the other? Or are the magnitudes equal? b. Compare the final speeds of the balls as they reach the ground. Is one larger than the other? Or are they equal?

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Here's an interesting challenge you can give to a friend. Hold a $1 (or larger) bill by an upper corner. Have a friend prepare to pinch a lower corner, putting her fingers near but not touching the bill. Tell her to try to catch the bill when you drop it by simply closing her fingers. This seems like it should be easy, but it's not. After she sees that you have released the bill, it will take her about 0.25 s to react and close her fingers-which is not fast enough to catch the bill. How much time does it take for the bill to fall beyond her grasp? The length of a bill is 16 cm.
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