Velocity and Acceleration of motorcycle

  • Thread starter sasspup29
  • Start date
  • #1
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Superguy has attached a nitro booster to his motorcycle to assist him in his quest to rid Science City of his archenemy, Mad Maxwell. The motorcycle doubles its acceleration when the nitro is activated. Superguy is idling by the side of the highway when Maxwell tears by. At this instant Superguy steps on the gas and accelerates at a constant rate until he is moving at the same speed as Maxwell, then uses his booster to close the distance and catch Maxwell. Maxwell moves at a constant 30.0 m/s the entire time, and is caught 60.0 seconds after passing Superguy. (a) How far does Maxwell travel after passing Superguy before being caught? (b) What is the acceleration of the motorcycle with the nitro booster engaged? (c) How long is Superguy in motion before engaging the booster?

Motion Equations for Constant Acceleration
1. Vf=Vi+at
2. X=vit+.5at2
3. Vf2=Vi2+2aX
4. X=.5(Vi+Vf)t

X=displacement
Vf= final velocity

For part a, I got 1800 m by using eq. 4. But I can't figure out part b using any of the equations because I don't know at what time he used the nitro booster or what the acceleration is. I think if I get a hint on part b I can figure out part c.

Thank you,
Sasspup29
 

Answers and Replies

  • #2
Delphi51
Homework Helper
3,407
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Welcome to PF, sasspup!
Your part (b) looks tough! I suggest you call that time when S catches up with M time 1 or t1. Then the time for the second part of the motion at double the acceleration is (60-t1). You'll have two unknowns (acceleration and t1) so you'll have to use two of your equations to get a system of equations you can solve for the two unknowns. Have a go!
 
  • #3
I solved this in a hurry and I haven't got integer answers (perhaps i made an error in simplification), but approximately, ans to b is 1.7 m/s2 and ans to c is about 35 seconds

you just have to apply those equations....
let the variables be a => acceleration without the nitro boost
and t => time till nitro boost is activated
d1=distance before nitro boost
d2=distance after nitro boost

then,
30=at from eqn 1

d1=at2/2 from 2

d2=30(60-t) +a(60-t)2 from 2

d1+d2=1800

solve for a and t
 
  • #4
2
0
Thank you both! I'll definitely be working this problem until I KNOW how you got the answers.
 

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