## Solution:

### Part A

From Newton’s second law of motion, we know that

$\text{net Force}=\text{mass}\times \text{acceleration}$

For this particular problem, we are solving for the mass. Hence, the formula for mass is

$\displaystyle \text{mass}=\frac{\text{net force}}{\text{acceleration}}$

In symbols, that is

$\displaystyle \text{m}=\frac{\sum \text{F}}{\text{a}}$

Substituting the given values and solving for the unknown, we have

$\displaystyle \text{m}=\frac{50.0\:\text{N}}{0.893\:\text{m/s}^2}$

$\text{m}=55.9910\:\text{kg}$

The mass of the austronaut is about 55.9910 kilograms.

## Part B

If the force could be exerted on the astronaut by another source other than the spaceship, then the spaceship would not experience a recoil.

## Solution:

From Newton’s second law of motion, we know that

$\text{net Force}=\text{mass}\times \text{acceleration}$

For this particular problem, we are solving for the acceleration. Hence, the formula for acceleration is

$\displaystyle \text{acceleration}=\frac{\text{net force}}{\text{mass}}$

In symbols, that is

$\displaystyle \text{a}=\frac{\sum \text{F}}{\text{m}}$

Substituting the given values and solving for the unknown, we have

$\displaystyle \text{a}=\frac{60.0\:\text{N}}{4.50\:\text{kg}}$

$\text{a}=13.33\:\text{m/s}^2$

The acceleration of the laundry cart is about 13.33 m/s2.

## Solution:

The velocity at time 2.5 seconds is 5 m/s.

The velocity at time 7.5 seconds is -4 m/s.

## Solution:

The position vs time graph is shown in the figure below.

## Solution:

Since the graph is a straight line, we can use the two points before and after the specified time to determine the slope of the line. The slope of the velocity-time graph is the acceleration.

The two points are $\left(0,\:165\right)\:and\:\left(20,\:228\right)$

The velocityis computed as

$a=m=\frac{\Delta y}{\Delta x}$

$a=\frac{y_2-y_2}{x_2-x_1}$

$a=\frac{228\:m/s-165\:m/s}{20\:s-0\:s}$

$a=3.2\:m/s^2$

Therefore, the acceleration is verified to be 3.2 m/s² at t=10 s.

## Solution:

Since the graph is a straight line, we can use any two points to determine the slope of the line. The slope of the position-time graph is the velocity.

The two points are $\left(20,\:6.95\right)\:and\:\left(40,\:11.7\right)$

The velocityis computed as

$v=m=\frac{\Delta y}{\Delta x}$

$v=\frac{y_2-y_2}{x_2-x_1}$

$v=\frac{11.7\:km-6.95\:km}{40\:s-20\:s}$

$v=0.238\:km/s$

Therefore, the velocity is 0.238 km/s.

## Solution:

Since the graph is a straight line, we can use any two points to determine the slope of the line. The slope of the position-time graph is the velocity.

The two points are $\left(0,\:2.80\right)\:and\:\left(20,\:6.95\right)$

The velocityis computed as

$v=m=\frac{\Delta y}{\Delta x}$

$v=\frac{y_2-y_2}{x_2-x_1}$

$v=\frac{6.95\:km-2.80\:km}{20\:s-0\:s}$

$v=0.208\:km/s$

Therefore, the velocity is 0.208 km/s.

## Solution:

### Part A

Based from the graph shown in figure 2.60, we can choose two points, one before and one after t=20 s. We have the points $\left(15,\:988\right)\:and\:\left(25,\:2138\right)$.

The slope is computed using the formula:

$m=\frac{\Delta y}{\Delta x}$

$m=\frac{2138\:m-988\:m}{25\:s-15\:s}$

$m=115\:m/s$

Therefore, we have verified that the velocity of the jet car is 115 m/s at t=20 seconds.

### Part B

Since the graph is a straight line, we can use any two points to determine the slope of the line. The slope of the velocity-time graph is the acceleration.

The two points are $\left(10,\:65\right)\:and\:\left(25,\:140\right)$

The acceleration is computed as

$a=m=\frac{\Delta y}{\Delta x}$

$a=\frac{y_2-y_2}{x_2-x_1}$

$a=\frac{140\:m/s-65\:m/s}{25\:s-10\:s}$

$a=5\:m/s^2$

Therefore, the acceleration is verified to be 5.0 m/s².