Find the magnitude of a particle's acceleration vector at t = 1 for a particle moving with position vector r(t)=(2t^2,4t^2-t^3)
a) 4
b) 25
c) 5
d) 2√5
r(t) = (2t^2,4t^2-t^3)
v(t) = dr/dt = (4t,8t-3t^2)
a(t) = dv/dt = (4,8-6t)
at t=1,
a(1) = (4,2)
|a(1)| = √(4^2+2^2) = 3√2
Hmm. I suspect a typo or a wrong answer key.
A typo indeed.
|a(1)| = √(4^2+2^2) = √20 = 2√5
I forgot to square the 2!
To find the magnitude of the particle's acceleration vector at t = 1, we need to find the second derivative of the position vector and evaluate it at t = 1.
First, let's find the first derivative of the position vector r(t):
r'(t) = (d/dt)(2t^2, 4t^2-t^3)
= (4t, 8t - 3t^2)
Next, let's find the second derivative of the position vector r(t):
r''(t) = (d/dt)(4t, 8t - 3t^2)
= (4, 8 - 6t)
Now, let's evaluate r''(t) at t = 1:
r''(1) = (4, 8 - 6(1))
= (4, 8 - 6)
= (4, 2)
The acceleration vector at t = 1 is (4, 2).
To find the magnitude of this vector, we use the formula:
Magnitude = sqrt((ax)^2 + (ay)^2)
Magnitude = sqrt((4)^2 + (2)^2)
= sqrt(16 + 4)
= sqrt(20)
= 2√5
Therefore, the magnitude of the particle's acceleration vector at t = 1 is 2√5.
The correct option is d) 2√5.
To find the magnitude of a particle's acceleration vector at a specific time t, we need to compute the second derivative of the position vector with respect to t and evaluate it at that time.
Given that the position vector of the particle is r(t) = (2t^2, 4t^2 - t^3), let's find its acceleration vector.
Step 1: Take the first derivative of the position vector with respect to t to find the velocity vector.
First derivative of r(t) = [(d/dt)(2t^2), (d/dt)(4t^2 - t^3)]
= (4t, 8t - 3t^2).
Step 2: Take the second derivative of the position vector with respect to t to find the acceleration vector.
Second derivative of r(t) = [(d/dt)(4t), (d/dt)(8t - 3t^2)]
= (4, 8 - 6t).
Now, we have the acceleration vector as a function of t, which is (4, 8 - 6t).
To find the magnitude of the acceleration vector at t = 1, substitute t = 1 into the acceleration vector.
Acceleration at t = 1 = (4, 8 - 6(1))
= (4, 2).
Finally, to find the magnitude of the acceleration vector, we use the Pythagorean theorem:
Magnitude of the acceleration = |(4, 2)|
= √(4^2 + 2^2)
= √(16 + 4)
= √20
= 2√5.
Therefore, the magnitude of the particle's acceleration vector at t = 1 is 2√5.
Hence, the correct answer is d) 2√5.