Velocity and Acceleration of motorcycle

[sasspup29]

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. V_f=V_i+at
2. X=v_it+.5at²
3. V_f²=V_i²+2aX
4. X=.5(V_i+V_f)t

X=displacement
V_f= 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

 

[Delphi51]

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!

 

[rishicomplex]

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

 

[sasspup29]

Thank you both! I'll definitely be working this problem until I KNOW how you got the answers.

 

[rwisz]

Hello Sasspup29,

I can help you with your questions about the velocity and acceleration of Superguy's motorcycle. Let's take a closer look at the information provided and use some physics principles to solve the problem.

First, we know that Superguy's motorcycle doubles its acceleration when the nitro booster is activated. This means that the acceleration of the motorcycle without the booster is half of the acceleration with the booster. Let's call this acceleration without the booster "a" and acceleration with the booster "2a".

Next, we know that Superguy is idling by the side of the highway when Maxwell passes by at a constant speed of 30.0 m/s. This means that at the moment Superguy starts accelerating, Maxwell is already 30.0 m/s ahead of him.

Now, let's look at the equations you mentioned. In order to solve for the distance that Maxwell travels before being caught, we can use equation 4, which is:

X = .5(Vi + Vf)t

We know that Vi (initial velocity) for Maxwell is 30.0 m/s, and Vf (final velocity) is also 30.0 m/s since he is moving at a constant speed. We also know that t (time) is 60.0 seconds, as stated in the problem. Plugging these values into the equation, we get:

X = .5(30.0 + 30.0)(60.0)
X = 1800 m

Therefore, Maxwell travels 1800 m before being caught by Superguy.

For part b, we can use equation 1, which is:

Vf = Vi + at

We know that Vi (initial velocity) for Superguy is 0 m/s, and Vf (final velocity) is also 30.0 m/s since he catches up to Maxwell. We also know that t (time) is the same for both Superguy and Maxwell, which is 60.0 seconds. Plugging these values into the equation, we get:

30.0 = 0 + a(60.0)
a = 0.5 m/s^2

Since the acceleration of the motorcycle without the booster is half of the acceleration with the booster, the acceleration with the booster is 1 m/s^2.

For part c, we can use equation 2, which is:

X =