# Physics homework: acceleration, velocity and speed. help please

1. Sep 20, 2014

### kirsten_2009

• Warning! Posting template must be used for homework questions; One question per thread.
Hello everyone,

I have this physics assignment to do and I would sure appreciate your guy's input. My attempts to answer the questions are below. Could you please let me know if I'm correct or if I'm wrong or missing anything? Thanks a lot in advance!

1. When you drive a car, might you depress the accelerator pedal without actually accelerating? Could you accelerate without having your foot on the accelerator? Explain.

No. If you press the accelerator then you are increasing the speed which affects velocity and thus you cannot affect velocity without causing acceleration.

Yes. By not pressing on the accelerator; you would assume that the car would slow down if going up a hill, or going on a flat surface (due to friction, gravity and air resistance) or speed up if going downhill and so a change in speed would occur, which would change the velocity, which in turn, would change the acceleration.

2. You drop a rock down a well and hear a splash 3 s later. How deep is the well assuming that air resistance is negligible and the time needed for sound to travel up the well is also negligible?

1/2 (9.8 m/s^2) x 3s^2 = 44.1m ***On a side note, why is there a 1/2 in front of gravity constant?

3. One car goes from 0 to 30 km/h. Later another car goes from 0 to 60 km/h. Can you say which car had greater acceleration? Explain.

I'm tempted to say that the 60km/h car underwent greater acceleration but I'm also tempted to say that you can't really tell because I don't know how long it took these cars to get to the speed they did. Could it not be the case that they accelerated at the same rate but one stopped speeding up and the other continued speeding up but at the same rate?

4. The acceleration due to gravity on the Moon is about 1/6 that on Earth. If an object is dropped near the surface of the Moon, how far does it fall in 10 s?

1/2 (1.63 m/s^2) x 10s^2 = 132 m.....? Doesn't look right....?

5. Charlie walks 1 km to the football stadium in 30 minutes, then sits and watches the game for 2 hours, then walks 1 km home in 30 minutes. What was Charlie’s average speed while he was gone?

2km/3 hr = 0.67 km/hr

2. Sep 20, 2014

### Ritzycat

I'll answer with what I am able to.

Number two is correct: 41m

As for number 3, you are correct on your second assumption. Even if you know initial and final velocities, you can't tell its acceleration without knowing the time it took to go from initial to final velocity. If they took the same time to get to their respective final velocities, then the 60km/hr car would have a greater velocity, but you can't say that because you don't know the time.

On number 4, I got 81.5m. I think what you did is this:

1/2(1.63m/s^2 * 10s)^2 = 132m
what shold be done is this:
1/2(1.63m/s^2)(10s)^2 = 81.5m

With this kinematic equation, make sure you are only squaring the time value. You squared the product of acceleration and time.

3. Sep 21, 2014

### kirsten_2009

Hi Ritzycat,

Thanks so much for taking the time to help me out. #4 makes sense now!

4. Sep 21, 2014

### Simon Bridge

#1 is incorrect.
Your two answers contradict each other - if you could slow down going up a hill without pressing the accelerator, then couldn;t you press the accelerator, while going up a hill, to maintain a constant speed?

#2 you should include your reasoning - why pick that calculation?
How did you account for the amount of time the sound took to get back to the top of the well?
Note: displacement is the area under the velocity-time graph - for the falling stone, what shape is the v-t graph?

#3 good thinking - without more information, you cannot tell. You need either the distance or the time.

#4 Try to articulate your doubts: why does that not look right to you?
This is the first step to trouble-shooting.

Why did you choose that equation?
How far would an object fall in 10secs on the Earth?

#5 Well done - C was gone for 3 hours and he covered a total distance of 2km.
Bonus points for not converting to m/s. It is usually OK just to say (2/3)kmph - you don't have to convert everything to decimal.
If you do convert to decimal - watch the sig fig matches up.

5. Sep 21, 2014

### kirsten_2009

Hello Mr. Bridge,

Thanks a lot for taking a look at my assignment! It absolutely makes sense. I'm looking into the v-t graph for #2 right now.

6. Sep 21, 2014

### Simon Bridge

Lets see - taking positive direction = "down": the stone starts at u=0 and accelerates uniformly to some speed v > 0 in some time t > 0.
So draw your axis. Mark a point on the time axis and label it t. You don't need to know the value of t here, any point so long as it is bigger than zero. Do the same for the velocity axis - mark it v. This is called "sketching" the graph.
Draw a line from (0,0) to (t,v).
Draw a dotted line from (t,v) to (t,0) ... what shape do you see?
What is the formula for the area of that shape?

If it takes time t for the stone to hit the water and time T to hear the sound, then t<T.
... and it takes the sound T-t time to get to the top.
The time T is what you are given, not t.

If the well isn't very deep, then the correction for the time the sound takes will be very small.

7. Sep 25, 2014

### kirsten_2009

Hello Mr. Bridge,

Sorry for my delay. I was swamped with homework; but I did what you suggested and plotted a t-v graph and got a straight line and the area under
the straight line is a triangle. The formula for the area of a triangle is height x base /2 so I guess the 1/2 comes in to account for the fact that it's not the area of a rectangle that we're looking for. This leads me to another question though....isn't the relationship between acceleration and time squared? So...shouldn't I be getting a half parabola? Thanks for all your time and help!

-K.

8. Sep 25, 2014

### Simon Bridge

The acceleration is the slope of the graph you just drew - so you can immediately find the relationship between the acceleration, the time, and the final velocity.
You have just worked out a relationship between displacement, time and the final velocity.
Now use these two simultaneous equations to find the relationship between displacement, time, and acceleration.

9. Sep 30, 2014

### SSWheels

Consider these two equations:
$$D=\frac{V_i + V_f}{2} \cdot t$$and
$$V_f = V_i + at$$
Substitute the 2nd equation into the 1st and see what you get...