1. ## Beginner Velocity/Acceleration question

Well, 2 actually.

"In stopping from an initial velocity of 35 m/s a motorcycle leaves a streak on the highway. Assuming a deceleration of 20 m/s^2 what was the distance in KM of the streak?"

Okay, so what I (think) I've got here is:

Initial Velocity = 35 m/s
Acceleration = -20 m/s^2
D= ?
Final Velocity = 0 m/s

What I don't know is how to crunch it properly, and the acceleration being in m/s^2 compounds the problem.

I feel silly asking, but any help would be tremendously appreciated.

2. Originally Posted by Caturdayz
Well, 2 actually.

"In stopping from an initial velocity of 35 m/s a motorcycle leaves a streak on the highway. Assuming a deceleration of 20 m/s^2 what was the distance in KM of the streak?"

Okay, so what I (think) I've got here is:

Initial Velocity = 35 m/s
Acceleration = -20 m/s^2
D= ?
Final Velocity = 0 m/s

What I don't know is how to crunch it properly, and the acceleration being in m/s^2 compounds the problem.

I feel silly asking, but any help would be tremendously appreciated.
Maybe this'll help

$t=\frac{v}{a}$ and $d=\frac{1}{2}at^2$

3. Originally Posted by Caturdayz
Well, 2 actually.

"In stopping from an initial velocity of 35 m/s a motorcycle leaves a streak on the highway. Assuming a deceleration of 20 m/s^2 what was the distance in KM of the streak?"

Okay, so what I (think) I've got here is:

Initial Velocity = 35 m/s
Acceleration = -20 m/s^2
D= ?
Final Velocity = 0 m/s

What I don't know is how to crunch it properly, and the acceleration being in m/s^2 compounds the problem.

I feel silly asking, but any help would be tremendously appreciated.
$v_f^2 = v_0^2 + 2a(\Delta x)$

solve for $\Delta x$ ... don't forget to convert to km.

4. Okay, so I took 35/-20 to get -1.75

I then took (1/2)(-20*-1.75)^(2)

Which equals 612.5

I then took the square root of 612.5, which came out to 24.75 km.

5. Using Skeeter's formula I got 0^2 = 35^2 + 2(-20)(deltaX)

Solving for (deltaX), where i got .0306 kilometers

6. Originally Posted by Caturdayz
Okay, so I took 35/-20 to get -1.75

I then took (1/2)(-20*-1.75)^(2)

Which equals 612.5

I then took the square root of 612.5, which came out to 24.75 km.
hmmm.... We've got a negative time here, and that's not good. Maybe...

$V=\int-20dt=-20t+v_0$ where $v_0$ is the initial velocity which we are given. So, we have

$V=-20t+35$

Now $s=\int(-20t+35)dt=-10t^2+35t+s_0$ where $s_0$ is the initial displacement.

At $t=0,s=0\Rightarrow{s}_0=0$

So we have $s=-10t^2+35t$

Now set V=0 to determine the time taken to stop, then sub this time into the displacement function to determine the distance taken to stop.

7. Originally Posted by Caturdayz
Using Skeeter's formula I got 0^2 = 35^2 + 2(-20)(deltaX)

Solving for (deltaX), where i got .0306 kilometers
and that would be correct.

8. Originally Posted by skeeter
and that would be correct.
Wow, I've really got to read things better...

"Beginner Velocity/Acceleration question"

I'm slow sometimes...