## Grantham PHY220 Week 1 Assignment Problem 8

### A stone is dropped from the roof of a high building. A second stone is dropped 1.25 s later. How long does it take for the stones to be 25.0 meters apart?

Solution:

Let t be the amount of time after the first stone is dropped. The distance from traveled by the first stone is

$y_1=\frac{1}{2}gt^2$

The distance traveled by the second stone is

$y_2=\frac{1}{2}g\left(t-1.25\right)^2$

The difference between the two stones is 25.0 m after time

$y_1-y_2=25.0$

$\frac{1}{2}\left(9.80\:m/s^2\right)t^2-\frac{1}{2}\left(9.80\:m/s^2\right)\left(t-1.25\right)^2=25$

$4.9t^2-4.9\left(t^2-2.5t+1.5625\right)=25.0$

$4.9t^2-4.9t^2+12.25t-7.65625=25.0$

$12.25t=25.0+7.65625$

$12.25t=32.65625$

$t=2.67\:s$

## Grantham PHY220 Week 1 Assignment Problem 7

### Explain a possible situation where you start with a positive velocity that decreases to a negative increasing velocity while there is a constant negative acceleration.

Solution:

An example of this situation is a free-fall. If an object is thrown upward, the initial velocity is positive. Then the velocity decreases until the object thrown will reach its maximum height and then it goes back with a negative increasing velocity. In this entire flight, the acceleration is a constant negative–the acceleration due to the Earth’s gravity.

## Grantham PHY220 Week 1 Assignment Problem 6

### A sports car moving at constant speed travels 150 m in 4.00 s. If it then brakes and comes to a stop while decelerating at a rate of 6.0 m/s2 , how long does it take to stop?

Solution:

The velocity before applying the brake is

$v=\frac{150\:m}{4.00\:s}=37.5\:m/s$

Solve for t until the velocity becomes zero

$v=v_0+at$

$t=\frac{v-v_0}{a}=\frac{0-37.5\:m/s}{-6.0\:m/s^2}=6.25\:s$

## Grantham PHY220 Week 1 Assignment Problem 5

### A car travels 120 meters in one direction in 20 seconds. Then the car returns ¾ of the way back in 10 seconds. Calculate the average speed of the car for the first part of the trip. Find the average velocity of the car.

Solution:

The average speed of the car for the first part of the trip is

$average\:speed=\frac{x}{t}=\frac{120\:m}{20\:s}=6\:m/s\:$

The average velocity of the car is

$\overline{v}=\frac{\Delta x}{\Delta t}=\frac{120-\frac{3}{4}\left(120\right)\:m}{20+10\:s}=\frac{30\:m}{30\:s}=1\:m/s$

## Grantham PHY220 Week 1 Assignment Problem 4

### What is larger 4000 L or $4\times 10^5\:cm^3$$4\times 10^5\:cm^3$  (and you must show your reasoning.)

Solution:

There are $1000\:or\:1\times 10^3\:cm^3$ in 1 L. So, in $4\times 10^5\:cm^3$ there are

$4\times 10^5\:cm^3\times \left(\frac{1\:L}{1\times 10^3\:cm^3}\right)=400\:L$

Therefore, the 4000 L is larger than $4\times 10^5\:cm^3$.

## Grantham PHY220 Week 1 Assignment Problem 3

### What is the surface area of a sphere of diameter $2.4\times 10^2$$2.4\times 10^2$ cm?

Solution:

$SA=4\pi r^2=4\pi \left(\frac{2.4\times 10^2\:cm}{2}\right)^2=1.8\times 10^5\:cm^2$

## Grantham PHY220 Week 1 Assignment Problem 2

#### Multiply $1.783\times 10^{-2}\cdot 4.4\times 10^{-3}$$1.783\times 10^{-2}\cdot 4.4\times 10^{-3}$, taking into account significant figures.

Solution:

$1.783\times 10^2\cdot 4.4\times 10^{-3}=0.78\:or\:7.8\times 10^{-1}$

## Grantham PHY220 Week 1 Assignment Problem 1

#### One lightyear is defined to be the distance light can travel in one year. What is this distance in meters? How long does it take for light to get to the Moon?

Solution:

$1\:lightyear=9.461\times 10^{15}\:m$

The approximate distance from Earth to the moon is 384,400 kilometers. The time it takes for the light to reach the moon is

$t=\frac{384400\:km\:\left(\frac{1000\:m}{1.00\:km}\right)}{299\:792\:458\:\frac{m}{s}}=1.28\:s$