Calculus: have distance and deceleration; need starting velocity.?

Question

Stopping distance: 60 meters (vehicle comes to complete stop)
Constant deceleration: 15 meters/sec^2

What was the starting velocity (before deceleration)?

I have a good understanding of derivatives and just learned antiderivatives. But, I don't know how to set problems like this up. Please help me understand how to set it up. I may ask a series of contrived questions like this.

Answer 1

With uniform deceleration f = 15 m/s^2

vf^2 = vi^2 + 2 ( -f) s
vf = 0 when stopped and s = 60m

vi^2 = 2 f s = 2 * 15 * 60 = 1800

vi = 42.42 m/sec
-------------------------
by calculus
equ of motion acceleration is
d^2x/dt^2 = dv/dt = d/dt [ dx/dt] = - 15

d [ dx/dt] = - 15 dt -----(a)

v = dx/dt

so dt = dx / v >>>> put in (a)

d [v ] = - 15 dx / v

v dv = - 15 dx
integrating for velocity

∫ v dv = - 15 ∫dx + c

[v^2/2] = - 15 x + c

given when x = 60, v = 0

0 = - 15 *60 + c >>> c = 900

[v^2/2] = - 15 x + 900

[v^2] = - 30 x + 1800

at x = 0, v = vi intitial

vi^2 = 1800

vi = 42.42 m/s

Answer 2

What we need is v(0), when t = 0
Start with deceleration, which is constant:

a = -15 m/s^2


Find "v" by integrating:

Integral [-15] dt = -15t + C, where C is the constant of integration


Find "s" by integrating again:

Integral [-15t + C] dt = (-15/2)t^2 + Ct + K, where K is the second constant of integration


We have:

v = -15t + C
s = (-15/2)t^2 + Ct + K

At t = 0, s = 0, (then when it is completely stopped, s can equal 60.)

0 = (-15/2)(0)^2 + C(0) + K, thus K = 0
and
s = (-15/2)t^2 + Ct

At s = 60, when the vehicle stops,
v = 0 = -15t + C ----> C = 15t


Also, at s = 60:
60 = (-15/2)t^2 + Ct ---> 60 = (-15/2)t^2 + 15t^2
60 = (15/2)t^2
t^2 = 8
t = 2sqrt(2)
C = 30sqrt(2)

v(0) = -15(0) + 30sqrt(2)

= 42 meters/second

Answer 3

Calculus will help you derive the basic equations of motion, but for problems this level of difficulty (e.g., not hard), you are probably better served understanding how calculus is used to get the equations of motion, then commit a few of the more common equations to memory.

For example, for linear motion, velocity is the integral of acceleration (where acceleration varies with time) over differential time:

v=∫a(t)dt, or v = v0 + at (if acceleration is constant).

Similarly, displacement is equal to velocity integrated over time. For constant acceleration, that is,

x=∫vdt = ∫(v0+at)dt = x0 + v0t + ½at²

Combining the first and second equations, you can remove the parametric time expression to get

v²=v0²+2a(x-x0)

So I've used calculus to show you how to get the equations of motion, but most of the time you won't need calculus to solve equations of motion (unless you're given an acceleration term that changes in time).

Which equations you use depends on what information you have and what you need. In the example you've given, you have distance and acceleration and you need velocity. So Equation 3 fits that build just fine:

v²=v0²+2a(x-x0)
0²=v0²+2(15)(-60)
v0² = 1800
v0=42.4m/s

Generally when you set up a problem, the most important things to understand are these. First, the term on the left is the "end state" of the motion (that's why v=0 instead of v0=0: v is the "end state", and we're told that the motion ends at zero. Second, even equations of motion in a line are vector quantities, which means sign matters. If you're traveling to the right, and you're decelerating, that means acceleration is pointing to the left and should be negative. Notice in your example I stuck in a minus sign in the expression; that's because acceleration in this instance is negative.

Same thing applies for throwing a ball in the air: the velocity might be upward (give it a positive sign), but acceleration due to gravity is downward (give it a negative sign).

So in short, almost all of linear kinematics can be reduced to the three equations above, and once you recognize what terms constitute your "initial conditions" (x0,v0), final state (the term to the left of the "=" sign), and whether the sign of a displacement, velocity, or acceleration is positive or negative, you will be able to set up any linear kinematics problem.

Good luck, work hard, and stay away from drugs.

Answer 4

Setting this up takes a trick in thinking:
a ≡ dv/dt = (dv/ds)(ds/dt) = vdv/ds (since v ≡ ds/dt)
∫ads = ∫vdv
a(s - s0) = (1/2) (v^2 - v0^2)
v^2 - v0^2 = 2a(s - s0)
v0^2 = v^2 - 2a(s - s0)
v0^2 = 0 - 2(-15)(60 - 0)
v0^2 = 30*60 = 1800
v0 = √1800 ≈ 42.426 m/s