Find the total distance traveled

[sweetdaisy186]

Hey guys!

I have a question on a second problem as well...

While chasing Mitch for destroying his car, Batman does the following in Skid's stolen Pinto:

1. He starts from rest and accelerates at 0.6 m/s^2 for 15 s
2. Travels at a constant velocity for 2 minutes
3. Slows to a stop with an acceleration of -0.75 m/s^2

What is the total distance that Batman travels?

I don't understand how I am supposed to approach this problem. Could someone please give me a hint so that I can figure it out on my own?

Thanks!

 

[Andrew Mason]

Hey guys!

I have a question on a second problem as well...

While chasing Mitch for destroying his car, Batman does the following in Skid's stolen Pinto:

1. He starts from rest and accelerates at 0.6 m/s^2 for 15 s
2. Travels at a constant velocity for 2 minutes
3. Slows to a stop with an acceleration of -0.75 m/s^2

What is the total distance that Batman travels?

I don't understand how I am supposed to approach this problem. Could someone please give me a hint so that I can figure it out on my own?

Thanks!

Draw a graph of velocity vs. time. What does the area under that graph mean physically? What is the area?

AM

 

[sweetdaisy186]

The graph shows that velocity can either increase or decrease over time depending on the slope. I'm confused. How does the graph help solve the problem?

 

[Hootenanny]

The graph shows that velocity can either increase or decrease over time depending on the slope. I'm confused. How does the graph help solve the problem?

I suggest you re-read AM's post, what does the area under the curve represent?

 

[HallsofIvy]

You should know:
change in velocity= acceleration*time
distance traveled= velocity* time.

1. He starts from rest and accelerates at 0.6 m/s^2 for 15 s

So how fast was he going at the end of the 15s?
What was his average speed during that time?
Hint: For a simple "linear" situation like this, with constant acceleration, the average speed is just the average of the beginning speed (which was 0) and the end speed.
How far would he go traveling at that average speed for 15 s?

2. Travels at a constant velocity for 2 minutes

And you just calculated what that constant velocity was, didn't you?
(NOT the average speed- his speed at the end of the 15 seconds.)
How far will he travel going at that speed for 2 min= 120 seconds?

3. Slows to a stop with an acceleration of -0.75 m/s^2[\quote]
How long does he have to de-cellerate at that amount in order to go from the constant speed you used above to 0?
What is his average speed during that time?
How far will he go at that average speed for the time you just calculated?

Now add those 3 distances to get the total distance.

 

[THE 1]

Making a graph is easiest but you could also work out each part using the kinematic equations if you know them a bit longer but its what works for you

 

[Andrew Mason]

The graph shows that velocity can either increase or decrease over time depending on the slope. I'm confused. How does the graph help solve the problem?

The area under the graph is equal to the distance travelled. v = ds/dt For a section where v does not change, $s = v\Delta t$. Where v is changing with time but the line is straight, the area is a triangle so $s = \frac{1}{2}v\Delta t$. You just have to work out the area under that graph.

AM

 

[sweetdaisy186]

ooooo. Okay, AM, we have not been taught the whole graph thing yet, that is why I was confused. We are only using equations right now. Thanks!

 

[sweetdaisy186]

For the time part, I got 13.5m for the change in velocity. For the second quote, I got 108 m because I multiplied the 0.9 m/s that I got when I muliplied 0.6 and 15s. I multiplied the 0.9 m/s with the 120 seconds to get the 108 m. Do the first two parts sound okay. I want to make sure that they are right before I go on. Thanks!

 

[Andrew Mason]

For the time part, I got 13.5m for the change in velocity. For the second quote, I got 108 m because I multiplied the 0.9 m/s that I got when I muliplied 0.6 and 15s. I multiplied the 0.9 m/s with the 120 seconds to get the 108 m. Do the first two parts sound okay. I want to make sure that they are right before I go on. Thanks!

Careful. The change in velocity is not in metres. And .6 * 15 = 9 m/s. not .9 m/s.

For each section, use:

$$d = v_0t + \frac{1}{2}at^2$$ and

$$v = v_0 + at$$

to find the position and velocity at time t. v0 is the speed at the beginning of the section and t is the time elapsed from the beginning of that section.

So for the first section, v0=0, a=.6m/s and t = 15 s, d = 0 +.5*.6*15^2 = 67.5 m. and v = 9 m/sec.

Use those values for the next sections to determine the distance and speed at the end.

AM

 

[sweetdaisy186]

Okay, so in the first part you have to find the distance that we went and at what velocity which is not the same thing as speed?. If you are driving at a constant velocity that means that the accerlation is zero right? To find the distance for the second part I have to multiply the velocity and time correct?

 

[Andrew Mason]

Okay, so in the first part you have to find the distance that we went and at what velocity which is not the same thing as speed?. If you are driving at a constant velocity that means that the accerlation is zero right? To find the distance for the second part I have to multiply the velocity and time correct?

Of course, velocity is speed and direction. In this case, there is no change in direction, so velocities and speeds are interchangeable. Just use the general expression for distance which I gave you as I have shown. For the second part s = vt where v is the speed at the end of the first part and t is the time (2 minutes). This is because a = 0 in the expression for distance.

You will notice that Skid's Pinto is not the Batmobile.

AM

 

[sweetdaisy186]

OOO I see! Thanks sooo much!