A particle is moving with the given data. Find the position of the particle.
a(t)= (t^2 - 5t +7) , s(0)=0 , s(1)=20
Thanks!
a = t^2-5t+7
v = 1/3 t^3 - 5/2 t^2 + 7t + C1
s = 1/12 t^4 - 5/6 t^3 + 7/2 t^2 + C1*t + C2
plugging s(0) and s(1), we have
C1 = 0
1/12 - 5/6 + 7/2 + 0 + C2 = 20
C2 = 69/4
so,
s(t) = 1/12 t^4 - 5/6 t^3 + 7/2 t^2 + 69/4
When I put your answer into WebAssign it marked as wrong so are you sure this is the right answer?
well, pretty sure. Do you see any errors in my work?
wolframalpha agrees with me.
http://www.wolframalpha.com/input/?i=solve+y%22+%3D+x^2-5x%2B7+where+y%280%29%3D0+and+y%281%29%3D20
actually, I do see a mistake.
C2 = 0
C1 = 69/4
s(t) = 1/12 t^4 - 5/6 t^3 + 7/2 t^2 + 69/4 t
To find the position of the particle, we need to integrate the acceleration function, a(t), twice with respect to time, t.
First, we need to find the velocity function, v(t), by integrating a(t) once. The integral of (t^2 - 5t + 7) with respect to t is (1/3)t^3 - (5/2)t^2 + 7t + C1, where C1 is the constant of integration.
Since the particle has an initial velocity of 0 (v(0) = 0), we can solve for C1:
v(0) = (1/3)(0)^3 - (5/2)(0)^2 + 7(0) + C1 = C1
C1 = 0
Therefore, the velocity function is v(t) = (1/3)t^3 - (5/2)t^2 + 7t.
Next, we need to find the position function, s(t), by integrating v(t) once again. The integral of (1/3)t^3 - (5/2)t^2 + 7t with respect to t is (1/12)t^4 - (5/6)t^3 + (7/2)t^2 + C2, where C2 is the constant of integration.
Since the particle has an initial position of 0 (s(0) = 0), we can solve for C2:
s(0) = (1/12)(0)^4 - (5/6)(0)^3 + (7/2)(0)^2 + C2 = C2
C2 = 0
Therefore, the position function is s(t) = (1/12)t^4 - (5/6)t^3 + (7/2)t.
Now, we can find the position of the particle at t = 1 by substituting t = 1 into the position function:
s(1) = (1/12)(1)^4 - (5/6)(1)^3 + (7/2)(1)
s(1) = (1/12) - (5/6) + 7/2
s(1) = 20
Therefore, the position of the particle at t = 1 is 20 units.