If the velocity is given as v=t^2-2t+2 what is the acceleration when v=10m/s?
a = dv/dt = 2 t - 2
if you really mean when v = 2 we have to go back and find the times when v = 2
2 = t^2 - 2 t + 2
t^2 - 2 t = 0
t(t-2) = 0
t = 0 or t = 2
so
a = 2 t - 2
= -2
or
= 2
I mean when(2t-2) when v = 10
10 = t^2 -2t + 2
t^2 - 2 t - 8 = 0
(t+2)(t-4) = 0
use + t
so t = 4
a = 2 t - 2 = 6
To find the acceleration when the velocity is given, we need to take the derivative of the velocity equation with respect to time.
Given: v = t^2 - 2t + 2
Step 1: Differentiate the velocity equation with respect to time (t) to find the acceleration (a):
a = dV/dt
Step 2: Take the derivative of v with respect to t:
Differentiating t^2, we get 2t.
Differentiating -2t, we get -2.
Differentiating 2, we get 0, as it is a constant.
So, a = 2t - 2.
Step 3: Substitute the given velocity value (v = 10 m/s) into the acceleration equation to find the value of t:
10 = 2t - 2
Step 4: Solve for t. Move -2 to the other side of the equation:
2t = 10 + 2
2t = 12
Divide both sides by 2:
t = 6
Step 5: Substitute the value of t (t = 6) back into the acceleration equation to find the acceleration (a):
a = 2(6) - 2
a = 12 - 2
a = 10 m/s^2
Therefore, the acceleration when v = 10 m/s is 10 m/s^2.