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Help with physics

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Thread replies: 11
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A car is driving at a constant speed when the driver applies the brakes, giving the car a deceleration of 3.50m/s^2. The car comes to a stop at a distance of 34.0m. What was the car's speed when it travelled 17.0m from the point where the brakes were applied?

1. 10.9m/s
2. 14,5m/s
3. 10,7m/s
4. 21,0m/s
5. 15.3m/s

How do I get the answer?
>>
>>5636268
I'M THE TODDLER
>>
>>5636268
Fug I can't remember physics lessons at all
>>
Use Kinematic equations for the x direction. Since u have a constant velocity
>>
I dunno, I'm bad at physics :(
>>
The equation for this would be "v_final^2 = v_initial^2 + 2*a*d". I forget how you derive this, but it follows from the others IIRC, so just consider it known a priori. Anyway, you need to use this twice: once with a final v of 0 and a distance of 34 to figure out your starting velocity, and then use that as the initial velocity with a distance of 17.
>>
>>5636302
>>5636321
>kinematic equations
holy shit that takes me back.
>>
>>5636268
Just remeber that acceleration is a second derivative of speed! That d = v*t + 0.5*a*t^2!
>>
It's 1. 10.9m/s
v0 is 15.43m/s.
>>
>>5636268
Some decimals are (,) - fuck yeah, Europe!

Using >>5636321's formula:
Full distance to calculated initial speed, v0:
0^2 = v0^2 + 2 * (-3.5) * 34
v0^2 = 238 [m^2/s^2]

Let's use this in the same formula for 17m distance:
v1^2 = 238 + 2 * (-3.5) * 17
v1 = sqrt(238 - 119) = 10.9

Amirite?
>>
>>5636403
... I love you
Thread posts: 11
Thread images: 1


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