
#1
Aug2906, 11:07 PM

P: 32

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! 



#2
Aug2906, 11:29 PM

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AM 



#3
Aug3006, 08:39 AM

P: 32

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?




#4
Aug3006, 08:47 AM

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PF Gold
P: 9,789

Find the total distance traveled 



#5
Aug3006, 09:06 AM

Math
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Thanks
PF Gold
P: 38,881

You should know:
change in velocity= acceleration*time distance traveled= velocity* time. 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? (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? [quote]3. Slows to a stop with an acceleration of 0.75 m/s^2[\quote] How long does he have to decellerate 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. 



#6
Aug3006, 09:07 AM

P: 20

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




#7
Aug3006, 10:12 AM

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AM 



#8
Aug3006, 04:36 PM

P: 32

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!




#9
Aug3006, 04:41 PM

P: 32

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!




#10
Aug3006, 05:37 PM

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For each section, use: [tex]d = v_0t + \frac{1}{2}at^2[/tex] and [tex]v = v_0 + at[/tex] 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 



#11
Aug3006, 07:13 PM

P: 32

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?




#12
Aug3006, 10:30 PM

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



#13
Aug3106, 09:50 PM

P: 32

OOO I see! Thanks sooo much!



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