t is in seconds and the coefficients have the proper units for the vector r to be in meters.
1. Find velocity vector v(t) and acceleration vector a(t) of the particle.
Answer: v(t)= 3.0 i+4.0t j
a(t)=4.0 j
2. Find the magnitude and direction of v(t) at t=1.0second
Answer: 5.0 m/s
theta=53 degrees with xaxis
I want ALL the steps.Please.
Thanks
Vectors problem.Involves calculus. The position of a particle is given by r = 3.0t i+ 2.0t^2 j+ 5.0 k where i,j,k are the component vectors?
2016-02-11 11:49 am
回答 (1)
2016-02-11 11:52 am
The derivative of 3t is 3.
The derivative of 2t^2 is 4t
The derivative of 5 is 0.
So the derivative of (3t, 2t^2, 5) is just the vector of their derivatives, (3, 4t, 0) or 3i + 4t*j. That's the velocity. The acceleration is the derivative of that. You know what the derivative of 3 and of 4t are, right?
2. t = 1 second means replace the variable "t" by the number "1". So the velocity is 3i + 4*1*j. And you know how to find the magnitude and direction of a vector given its components, right?
The derivative of 2t^2 is 4t
The derivative of 5 is 0.
So the derivative of (3t, 2t^2, 5) is just the vector of their derivatives, (3, 4t, 0) or 3i + 4t*j. That's the velocity. The acceleration is the derivative of that. You know what the derivative of 3 and of 4t are, right?
2. t = 1 second means replace the variable "t" by the number "1". So the velocity is 3i + 4*1*j. And you know how to find the magnitude and direction of a vector given its components, right?
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