I'm having some trouble with this problem:
A sprinter explodes out of the starting block with an acceleration of +2.3 m/s^2, which she sustains for 1.2 s. Then, her acceleration drops to zero for the rest of the race. What is her velocity (a) at t= 1.2 s and (b) at the end of the race.
So, would the equation for (a) be (2.3) - (0)/1.2? And (b) would be 0/0?
Thanks in advance!
A sprinter explodes out of the starting block with an acceleration of +2.3 m/s^2, which she sustains for 1.2 s. Then, her acceleration drops to zero for the rest of the race. What is her velocity (a) at t= 1.2 s and (b) at the end of the race.
So, would the equation for (a) be (2.3) - (0)/1.2? And (b) would be 0/0?
Thanks in advance!
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Nope. Remember that whenever acceleration is constant:
v(t) = at + v0
...where we can assume v0 is zero. So when she's accelerating:
v(t) = 2.3t, t ≤ 1.2 s
At 1.2 acceleration drops to zero; this is where initial value comes into play.
v(t) = v(1.2), t ≥ 1.2 s.
At t = 1.2, here acceleration is zero. That means she will be running at the same velocity for the rest of the race, and that velocity is whatever her velocity was at 1.2 s from the first equation. So effectively, both (a) and (b) are asking for the same value.
v(1.2) = (2.3 m/s²)(1.2 s) = 2.76 m/s
v(end of race) = v(1.2) = 2.76 m/s
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EDIT: Oh, I think I see what you're going on about, but you're botching the equations. Typically, when dividing a difference by time, you're finding an average value, and since the acceleration is constant, the average of the velocities over time should be the constant acceleration. I suppose you could do:
[v(f) - v(i)]/t = a
...where v(f) is final velocity, v(i) is initial velocity, and a is a constant acceleration. So...
[v(1.2) - v(0)]/(1.2 s) = 2.3 m/s²
Since v(0) = 0, then...
[v(1.2) - 0]/(1.2 s) = 2.3 m/s²
v(1.2)/(1.2 s) - 0 = 2.3 m/s²
v(1.2) = (2.3 m/s²)(1.2 s) = 2.76 m/s
And then when acceleration is zero...
[v(f) - v(i)]/t = 0
v(f) - v(i) = 0
v(f) = v(i)
v(end of race) = v(1.2) = 2.76 m/s
I think that's what you're trying to do. If velocity is constant, then yes, you could use this same trick to find displacement. So you couldn't use [x(f) - x(i)]/t = v until her acceleration is zero, because before then, her velocity is changing.
v(t) = at + v0
...where we can assume v0 is zero. So when she's accelerating:
v(t) = 2.3t, t ≤ 1.2 s
At 1.2 acceleration drops to zero; this is where initial value comes into play.
v(t) = v(1.2), t ≥ 1.2 s.
At t = 1.2, here acceleration is zero. That means she will be running at the same velocity for the rest of the race, and that velocity is whatever her velocity was at 1.2 s from the first equation. So effectively, both (a) and (b) are asking for the same value.
v(1.2) = (2.3 m/s²)(1.2 s) = 2.76 m/s
v(end of race) = v(1.2) = 2.76 m/s
_______________________________________…
EDIT: Oh, I think I see what you're going on about, but you're botching the equations. Typically, when dividing a difference by time, you're finding an average value, and since the acceleration is constant, the average of the velocities over time should be the constant acceleration. I suppose you could do:
[v(f) - v(i)]/t = a
...where v(f) is final velocity, v(i) is initial velocity, and a is a constant acceleration. So...
[v(1.2) - v(0)]/(1.2 s) = 2.3 m/s²
Since v(0) = 0, then...
[v(1.2) - 0]/(1.2 s) = 2.3 m/s²
v(1.2)/(1.2 s) - 0 = 2.3 m/s²
v(1.2) = (2.3 m/s²)(1.2 s) = 2.76 m/s
And then when acceleration is zero...
[v(f) - v(i)]/t = 0
v(f) - v(i) = 0
v(f) = v(i)
v(end of race) = v(1.2) = 2.76 m/s
I think that's what you're trying to do. If velocity is constant, then yes, you could use this same trick to find displacement. So you couldn't use [x(f) - x(i)]/t = v until her acceleration is zero, because before then, her velocity is changing.