Question 1

A woman is running at 5.1 m/s to catch a bus parked at the bus stop. When she is 11 m from the bus, the bus leaves the stop with an acceleration of 1 m/s² (away from the woman). When the woman and the bus meet at the second instance (second location), how far away is she (or the bus) from the bus stop?

Accepted Answer

The answer of A woman is running at 5.1 m/s...

Question 2

The position of a particle is given by $x ( t ) = 23 t - 1.2 t ^ { 3 }$ , where x(t) is in meters and t is in seconds. The initial ( $t = 0$ E) position of the particle

Accepted Answer

The initial position of the particle is found by substituting $t = 0$ into the equation $x(t) = 23t - 1.2t^3$, which gives $x(0) = 23(0) - 1.2(0)^3 = 0$.

Question 3

The position of a particle is given by $x ( t ) = 23 t - 1.2 t ^ { 3 }$ , where x(t) is in meters and t is in seconds. The initial ( $t = 0$ ) velocity of the particle

Accepted Answer

The velocity of the particle is given by the derivative of the position function with respect to time, $v(t) = dx/dt$. For $x(t) = 23t - 1.2t^3$, $v(t) = d(23t - 1.2t^3)/dt = 23 - 3.6t^2$. At $t = 0$, $v(0) = 23 - 3.6(0)^2 = 23$ m/s.

Question 4

The position of a particle is given by $x ( t ) = 23 t - 1.2 t ^ { 3 }$ , where x(t) is in meters and t is in seconds. The initial ( $t = 0$ ) acceleration of the particle

Accepted Answer

The acceleration of the particle is found by taking the second derivative of the position function $x(t) = 23t - 1.2t^3$ with respect to time $t$. The first derivative $x'(t)$ gives the velocity, and the second derivative $x''(t)$ gives the acceleration. Differentiating $x(t)$ once gives $x'(t) = 23 - 3.6t^2$, and differentiating again gives $x''(t) = -7.2t$. Evaluating this at $t = 0$ gives $x''(0) = -7.2(0) = 0$, indicating that the initial acceleration is 0 m/s^2.

Question 5

The position of a particle is given by $x ( t ) = 23 t - 1.2 t ^ { 3 }$ , where x(t) is in meters and t is in seconds. The times ( $t \geq 0$ ) where the particle is located at the origin are

Accepted Answer

To find when the particle is at the origin, set $x(t) = 0$: $23t - 1.2t^3 = 0$. Factoring gives $t(23 - 1.2t^2) = 0$, so $t = 0$ or $23 = 1.2t^2$. Solving $1.2t^2 = 23$ gives $t \approx 4.4$. Thus, the particle is at the origin at $t = 0$ and $t \approx 4.4$ seconds.

Question 6

The position of a particle is given by $x ( t ) = 23 t - 1.2 t ^ { 3 }$ , where x(t) is in meters and t is in seconds. The (positive) time when the particle has zero velocity is

Accepted Answer

The velocity is the derivative of the position function: $v(t) = \frac{d}{dt}(23t - 1.2t^3) = 23 - 3.6t^2$. Setting $v(t) = 0$ gives $23 - 3.6t^2 = 0$. Solving for $t$, we get $t^2 = \frac{23}{3.6}$, so $t = \sqrt{\frac{23}{3.6}} \approx 2.5$ seconds.

Question 7

All of the following quantities include the concept of the passage of time except

Accepted Answer

Position is a description of an object's location and does not inherently involve the passage of time, unlike motion, velocity, and acceleration, which all describe changes in position over time.

Question 8

We will always calculate the average speed correctly by simply taking half the sum of the initial and final speeds during the time interval for all of the following cases except

Accepted Answer

The average speed is correctly calculated by taking the total distance traveled divided by the total time taken. The method of taking half the sum of the initial and final speeds only gives the correct average speed when the acceleration is constant (uniform acceleration). In cases where the acceleration is not constant, this method does not necessarily yield the correct average speed, as the speed could be changing in a non-linear manner throughout the time interval.

Question 9

The average velocity of a body over some time interval is equal to the instantaneous velocity of the body whenever the acceleration of the body during that time interval is

Accepted Answer

The average velocity equals the instantaneous velocity when the acceleration is zero, meaning the velocity is constant throughout the interval.

Question 10

The mathematical relationship between the acceleration and the speed of a body is precisely the same as the mathematical relationship between

Accepted Answer

Acceleration is the rate of change of speed with respect to time, just as speed is the rate of change of distance with respect to time.

Question 11

The slope of a graph plotting the velocity of a body versus its time of travel is equal to

Accepted Answer

The slope of a velocity vs. time graph represents the rate of change of velocity with respect to time, which is the definition of acceleration.

Question 12

If a body is falling at its terminal speed, in the next instant

Accepted Answer

When a body is falling at its terminal speed, it means that the force of gravity pulling it downwards is balanced by the air resistance pushing it upwards, resulting in no net force and thus no acceleration. The body moves at a constant speed.

Question 13

All of the following are valid (though perhaps unusual) units for speed except

Accepted Answer

Speed is defined as the rate at which an object covers distance. It is typically expressed as a distance divided by time (e.g., meters/second, miles/minute, inches/year). A light-year, however, is a unit of distance, not speed. It measures the distance light travels in a vacuum in one year.

Question 14

All of the following are valid (though perhaps unusual) units for acceleration except

Accepted Answer

Acceleration is defined as the rate of change of velocity over time, which means it should be measured in units of distance per time squared. Options A, B, and D all follow this format, but option C (centimeters/second) is a unit of speed, not acceleration, because it lacks the squared time component.

Question 15

All of the following are approximately correct (that is, are exact within 10%) for a body falling freely (from rest) near the surface of the Earth except

Accepted Answer

The acceleration due to gravity near the Earth's surface is approximately 9.8 meters per second squared, or about 32 feet per second squared. This means that for each second of free fall, the speed of the object increases by about 32 feet per second, not the distance fallen. After one second, the body would have fallen only 16 feet (using the formula distance = 1/2 * g * t^2, where g is the acceleration due to gravity and t is the time in seconds), making option C incorrect.

Question 16

Dimensionally, correct relations among time [t], distance [x], velocity [v], and acceleration [a] include all of the following except

Accepted Answer

DOption A is incorrect because acceleration ([a]) is equal to velocity ([v]) divided by time ([t]), not velocity times distance. The correct dimensional formula for acceleration is [LT^-2], not [LT^-1][L] as implied by option A. Option D is also dimensionally incorrect because it suggests velocity squared ([v^2]) is equal to acceleration times distance, which is dimensionally consistent ([LT^-2][L] = [L^2T^-2]), but it's not a direct relation among the fundamental quantities as asked.

Question 17

A tourist starts along a path of total length L. He finds that in half the time allotted he has covered a distance of only L/3. This gives him an average speed of V₁. To complete the tour on schedule, his average speed for the remainder of the tour must be

Accepted Answer

The average speed for the entire tour must be L/t, where t is the total time allotted. Since the tourist has covered L/3 in half the time, the average speed for the first half of the tour is V₁ = L/2t. Therefore, the average speed for the second half of the tour must be 2V₁ in order to complete the tour on schedule.

Question 18

A tourist starts along a path of total length L at an average speed of V₁. At the halfway point (in distance), he finds that he has used two-thirds of the time allotted for the tour. To complete the tour on schedule, his average speed for the remainder of the tour must be

Accepted Answer

The tourist has used two-thirds of the time allotted for the tour at the halfway point, so he has one-third of the time remaining. To complete the tour on schedule, he must travel the remaining distance in one-third of the time. This means that his average speed for the remainder of the tour must be three times his original average speed, or 2V1.

Question 19

For a falling body near the surface of the Earth (neglecting air drag), all of the following change with time except

Accepted Answer

The acceleration of a falling body near the Earth's surface, due to gravity, is constant and approximately 9.8 m/s^2, assuming air resistance is negligible.

Question 20

For a falling body near the surface of the Earth (explicitly including air drag), all of the following change with time except

Accepted Answer

The terminal speed of a body falling through a fluid (like air) is the maximum speed it will reach, where the force of gravity is balanced by the drag force. This speed does not change with time once it is reached, unlike acceleration, velocity, and position, which all vary over the course of the fall.

Quiz 2: Motion Along a Straight Line

The position of a particle is given by $x ( t ) = 23 t - 1.2 t ^ { 3 }$ , where x(t) is in meters and t is in seconds. The initial ( $t = 0$ E) position of the particle

The position of a particle is given by $x ( t ) = 23 t - 1.2 t ^ { 3 }$ , where x(t) is in meters and t is in seconds. The initial ( $t = 0$ ) velocity of the particle

The position of a particle is given by $x ( t ) = 23 t - 1.2 t ^ { 3 }$ , where x(t) is in meters and t is in seconds. The initial ( $t = 0$ ) acceleration of the particle

The position of a particle is given by $x ( t ) = 23 t - 1.2 t ^ { 3 }$ , where x(t) is in meters and t is in seconds. The times ( $t \geq 0$ ) where the particle is located at the origin are

The position of a particle is given by $x ( t ) = 23 t - 1.2 t ^ { 3 }$ , where x(t) is in meters and t is in seconds. The (positive) time when the particle has zero velocity is

All of the following quantities include the concept of the passage of time except

We will always calculate the average speed correctly by simply taking half the sum of the initial and final speeds during the time interval for all of the following cases except

The average velocity of a body over some time interval is equal to the instantaneous velocity of the body whenever the acceleration of the body during that time interval is

The mathematical relationship between the acceleration and the speed of a body is precisely the same as the mathematical relationship between

The slope of a graph plotting the velocity of a body versus its time of travel is equal to

If a body is falling at its terminal speed, in the next instant

All of the following are valid (though perhaps unusual) units for speed except

All of the following are valid (though perhaps unusual) units for acceleration except

All of the following are approximately correct (that is, are exact within 10%) for a body falling freely (from rest) near the surface of the Earth except

Dimensionally, correct relations among time [t], distance [x], velocity [v], and acceleration [a] include all of the following except

A tourist starts along a path of total length L. He finds that in half the time allotted he has covered a distance of only L/3. This gives him an average speed of V₁. To complete the tour on schedule, his average speed for the remainder of the tour must be

A tourist starts along a path of total length L at an average speed of V₁. At the halfway point (in distance) , he finds that he has used two-thirds of the time allotted for the tour. To complete the tour on schedule, his average speed for the remainder of the tour must be

For a falling body near the surface of the Earth (neglecting air drag) , all of the following change with time except

For a falling body near the surface of the Earth (explicitly including air drag) , all of the following change with time except

Quiz 2: Motion Along a Straight Line

The position of a particle is given by x(t) =23t−1.2t3x ( t ) = 23 t - 1.2 t ^ { 3 }x(t) =23t−1.2t3 , where x(t) is in meters and t is in seconds. The initial ( t=0t = 0t=0 E) position of the particle

The position of a particle is given by x(t) =23t−1.2t3x ( t ) = 23 t - 1.2 t ^ { 3 }x(t) =23t−1.2t3 , where x(t) is in meters and t is in seconds. The initial ( t=0t = 0t=0 ) velocity of the particle

The position of a particle is given by x(t) =23t−1.2t3x ( t ) = 23 t - 1.2 t ^ { 3 }x(t) =23t−1.2t3 , where x(t) is in meters and t is in seconds. The initial ( t=0t = 0t=0 ) acceleration of the particle

The position of a particle is given by x(t) =23t−1.2t3x ( t ) = 23 t - 1.2 t ^ { 3 }x(t) =23t−1.2t3 , where x(t) is in meters and t is in seconds. The times ( t≥0t \geq 0t≥0 ) where the particle is located at the origin are

The position of a particle is given by x(t) =23t−1.2t3x ( t ) = 23 t - 1.2 t ^ { 3 }x(t) =23t−1.2t3 , where x(t) is in meters and t is in seconds. The (positive) time when the particle has zero velocity is

All of the following quantities include the concept of the passage of time except

We will always calculate the average speed correctly by simply taking half the sum of the initial and final speeds during the time interval for all of the following cases except

The average velocity of a body over some time interval is equal to the instantaneous velocity of the body whenever the acceleration of the body during that time interval is

The mathematical relationship between the acceleration and the speed of a body is precisely the same as the mathematical relationship between

The slope of a graph plotting the velocity of a body versus its time of travel is equal to

If a body is falling at its terminal speed, in the next instant

All of the following are valid (though perhaps unusual) units for speed except

All of the following are valid (though perhaps unusual) units for acceleration except

All of the following are approximately correct (that is, are exact within 10%) for a body falling freely (from rest) near the surface of the Earth except

Dimensionally, correct relations among time [t], distance [x], velocity [v], and acceleration [a] include all of the following except

A tourist starts along a path of total length L. He finds that in half the time allotted he has covered a distance of only L/3. This gives him an average speed of V1. To complete the tour on schedule, his average speed for the remainder of the tour must be

A tourist starts along a path of total length L at an average speed of V1. At the halfway point (in distance) , he finds that he has used two-thirds of the time allotted for the tour. To complete the tour on schedule, his average speed for the remainder of the tour must be

For a falling body near the surface of the Earth (neglecting air drag) , all of the following change with time except

For a falling body near the surface of the Earth (explicitly including air drag) , all of the following change with time except

The position of a particle is given by $x ( t ) = 23 t - 1.2 t ^ { 3 }$ , where x(t) is in meters and t is in seconds. The initial ( $t = 0$ E) position of the particle

The position of a particle is given by $x ( t ) = 23 t - 1.2 t ^ { 3 }$ , where x(t) is in meters and t is in seconds. The initial ( $t = 0$ ) velocity of the particle

The position of a particle is given by $x ( t ) = 23 t - 1.2 t ^ { 3 }$ , where x(t) is in meters and t is in seconds. The initial ( $t = 0$ ) acceleration of the particle

The position of a particle is given by $x ( t ) = 23 t - 1.2 t ^ { 3 }$ , where x(t) is in meters and t is in seconds. The times ( $t \geq 0$ ) where the particle is located at the origin are

The position of a particle is given by $x ( t ) = 23 t - 1.2 t ^ { 3 }$ , where x(t) is in meters and t is in seconds. The (positive) time when the particle has zero velocity is

A tourist starts along a path of total length L. He finds that in half the time allotted he has covered a distance of only L/3. This gives him an average speed of V₁. To complete the tour on schedule, his average speed for the remainder of the tour must be

A tourist starts along a path of total length L at an average speed of V₁. At the halfway point (in distance) , he finds that he has used two-thirds of the time allotted for the tour. To complete the tour on schedule, his average speed for the remainder of the tour must be