Question

At a place where \(g=9.8 \frac{\mathrm{m}}{\mathrm{s}^{2}},\) an object is thrown vertically downward with a speed of \(5 \frac{\mathrm{m}}{\mathrm{s}}\) while a different object is thrown vertically upward with a speed of \(10 \frac{\mathrm{m}}{\mathrm{s}} .\) Which object undergoes a greater change in speed in a time of 2 s? (A) The first object has a greater change in speed. (B) The second object has a greater change in speed. (C) Both objects undergo the same change in speed. (D) It cannot be determined from the information given.

Short answer

Both objects undergo the same change in speed.

Step-by-step solution

Identify Given Information

The acceleration due to gravity is given as \(g = 9.8 \frac{\mathrm{m}}{\mathrm{s}^2}\). One object is thrown downward with an initial speed of \(5 \frac{\mathrm{m}}{\mathrm{s}}\), and another object is thrown upward with an initial speed of \(10 \frac{\mathrm{m}}{\mathrm{s}}\). The time is \(2\) seconds.

Calculate the Final Speed of the Downward-Thrown Object

For the object thrown downward, use the equation of motion: \(v = u + at\), where \(u = 5 \frac{\mathrm{m}}{\mathrm{s}}, a = 9.8 \frac{\mathrm{m}}{\mathrm{s}^2}\), and \(t = 2\) seconds. Plugging in the values, we get: \[ v = 5 + 9.8 \times 2 \] \[ v = 5 + 19.6 = 24.6 \frac{\mathrm{m}}{\mathrm{s}} \] The final speed is \(24.6 \frac{\mathrm{m}}{\mathrm{s}}\).

Calculate the Change in Speed for the Downward-Thrown Object

The speed change for the first object is: \[ \Delta v = v_{final} - v_{initial} \] \[ \Delta v = 24.6 - 5 = 19.6 \frac{\mathrm{m}}{\mathrm{s}} \] The change in speed for the downward-thrown object is \(19.6 \frac{\mathrm{m}}{\mathrm{s}}\).

Calculate the Final Speed of the Upward-Thrown Object

For the object thrown upward, use the same equation of motion: \(v = u + at\), but note the direction of acceleration due to gravity is opposite to the initial velocity, so \(a = -9.8 \frac{\mathrm{m}}{\mathrm{s}^2}\). \(u = 10 \frac{\mathrm{m}}{\mathrm{s}}\) and \(t = 2\) seconds. Plugging in the values, we get: \[ v = 10 + (-9.8) \times 2 \] \[ v = 10 - 19.6 = -9.6 \frac{\mathrm{m}}{\mathrm{s}} \] The final speed is \(9.6 \frac{\mathrm{m}}{\mathrm{s}}\) (magnitude).

Calculate the Change in Speed for the Upward-Thrown Object

The speed change for the second object is: \[ \Delta v = |v_{final}| + |v_{initial}| \] \[ \Delta v = 9.6 + 10 = 19.6 \frac{\mathrm{m}}{\mathrm{s}} \] The change in speed for the upward-thrown object is \(19.6 \frac{\mathrm{m}}{\mathrm{s}}\).

Compare the Changes in Speed

Both objects experience a change in speed of \(19.6 \frac{\mathrm{m}}{\mathrm{s}}\). Hence, both objects undergo the same change in speed.

Key concepts

acceleration due to gravity

In projectile motion, one crucial factor to consider is the acceleration due to gravity, often denoted as \(g\). On Earth, \(g\) typically equals \(9.8 \frac{\mathrm{m}}{\mathrm{s}^{2}}\). This acceleration acts downward, regardless of the object's initial motion direction. Gravity influences how an object's speed changes over time. For instance, if you drop an object, it accelerates towards the ground because gravity pulls it. Conversely, if you throw an object upwards, gravity slows it down until it stops momentarily and then accelerates it back down. In the given exercise, both objects are affected by the same gravitational pull of \(9.8 \frac{\mathrm{m}}{\mathrm{s}^{2}}\). This constant acceleration ensures they have comparable speed changes after 2 seconds.

kinematics equations

Kinematics equations describe the motion of objects under uniform acceleration. These equations enable us to compute the final velocity, displacement, and time based on given initial conditions. One essential equation used in the problem is \(v = u + at\), where \(v\) is the final velocity, \(u\) is the initial velocity, \(a\) is the acceleration, and \(t\) is the time. For the downward-thrown object: \(v = 5 + 9.8 \times 2 = 24.6 \frac{\mathrm{m}}{\mathrm{s}}\). For the upward-thrown object, considering gravity acts opposite to the motion direction: \(v = 10 + (-9.8) \times 2 = -9.6 \frac{\mathrm{m}}{\mathrm{s}}\). This equation helps predict how fast an object will move at any given time during its motion.

initial velocity

Initial velocity is the speed at which an object starts its motion. In this problem, one object is thrown downward with an initial velocity of \(5 \frac{\mathrm{m}}{\mathrm{s}}\), and another object is thrown upward with an initial velocity of \(10 \frac{\mathrm{m}}{\mathrm{s}}\). Initial velocity is crucial because it influences how the object's speed changes over time. For the downward-thrown object, the initial velocity works with gravity to increase its speed. In contrast, for the upward-thrown object, the initial velocity works against gravity, which slows it down until the object momentarily stops before changing direction. Thus, initial velocity combined with gravitational acceleration determines each object's final speed after a given time.

final velocity

Final velocity is the speed of an object at a specific time after it has been in motion. This value can be found using the kinematics equation \(v = u + at\). For the downward-thrown object in our problem, its final velocity after 2 seconds of falling is \(24.6 \frac{\mathrm{m}}{\mathrm{s}}\), calculated by adding the initial velocity (\'5 \frac{\mathrm{m}}{\mathrm{s}}\') to the product of gravity and time (\'9.8 \frac{\mathrm{m}}{\mathrm{s}^{2}} \times 2\'). For the upward-thrown object, the final velocity is \(-9.6 \frac{\mathrm{m}}{\mathrm{s}}\), indicating it was moving downward at this speed due to gravity overcoming the initial upward motion. Understanding final velocity helps in determining overall motion changes over any given period.

change in speed

Change in speed, or \(\Delta v\) is the difference between the final speed and the initial speed of an object. It shows how much the object's speed increases or decreases over time. We can calculate it using \(\Delta v = v_{final} - v_{initial}\). For the downward-thrown object, the change in speed after 2 seconds is \(\Delta v = 24.6 - 5 = 19.6 \frac{\mathrm{m}}{\mathrm{s}}\). For the upward-thrown object, the change is \(\Delta v = |-9.6 - 10| = 19.6 \frac{\mathrm{m}}{\mathrm{s}}\). Both objects show the same change in speed due to the same magnitude of gravitational acceleration acting on them, highlighting that gravity equally affects the upward and downward motion in terms of speed change.