The position vector of a particle is given by r(t)=t^3*i+t^2*j. What are it's velocity speed and acceleration when t=2
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We have the position vector given in terms of time t. r(t) = t^3*i + t^2*j
To find the velocity vector we have to differentiate r(t) with respect to time.
r'(t) = 3t^2*i + 2t*j
The vector representing acceleration is the derivative of the position vector
r''(t) = 6t*i + 2*j
When time t = 2.
The velocity vector is 3*2^2*i + 2*2*j
=> 12*i + 4*j
The speed is the absolute value of the velocity vector or sqrt(12^2 + 4^2) = sqrt (144 + 16) = sqrt 160
The acceleration vector is 6*2*i + 2*j
=> 12*i + 2*j
The required acceleration at t=2 is 12*i + 2*j and the speed is sqrt 160.
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The velocity is given by the 1st derivative of r(t).
We'll differentiate r(t) with respect to t.
v(t) = r'(t) = 3t^2*i+ 2t*j
The acceleration is given by the 1st derivative of v(t).
We'll differentiate v(t) with respect to t.
a(t) = v'(t) = 6t*i + 2j
The speed is:
|v(t)| = sqrt[(3t^2)^2 + (2t)^2]
|v(t)| = sqrt(9t^4 + 4t^2)
We'll put t = 2 and we'll get:
v(2) = 3*(2)^2*i+ 4*j
v(2) = 12i + 4j
a(2) = 12i + 2j
|v(2)| = sqrt(144+ 16)
|v(2)| = sqrt(160)
|v(2)| = 4sqrt10
The velocity and acceleration of the particle, when t=2, are: v(2) = 12i + 4j ; a(2) = 12i + 2j and |v(2)| = 4sqrt10.
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