Question

An object's position is given by the equation:

a. What is the velocity of the object as a function of time? 

b. What is the acceleration of the object as a function of time? 

c. What is the magnitude of the instantaneous velocity at t = 3.0 s? 

d. What is the direction of the acceleration? (specify an angle relative to the + x axis).

Answer 1

R(t) = (4t + 6t^2) i + (-4 + 1.5 t^2) j v (t) = dr/dt = (4 + 12t) i + 3t j = > (d/dt (4t) + d 6t^2/dt) i + (d (-4)/dt + d 1.5 t^2/dt) j = (4 + 2 times 6t) i + (0 + 2 times 1.5 t) j a = dv/dt = d (4 + 12t)i/dt + phi (3t)/dt j = 12i + 3j v (3) = (4 + 12 times 3) i + 3 times 3j = 40 i + 9j Vertical-bar V (3) Vertical-bar = squareroot (40)^2 + (9)^2 = squareroot 1681 = > m/s Tan theta = a_y/a_x theta = tan^-1 (3/12) theta^-1 (0.25) theta = 14.036 degree

Answer 2

Answer 3

Rate of change of position vector is velocity. And rate of change of velocity of an object is the acceleration.

Calculation and explanation is shown below.

Rate of change of position vector is velocity. And rate of change of velocity of an object is the acceleration. Calculation and explanation is shown below. Here object position is given in vector form. r^vector (t) = (4 t + 6t^2) i^^+ (-4 + 1.5 t^2) j^^m/s To find velocity of the object as a function of time As we know rate of change of position of an object is equal to the velocity. V^vector (t) = dr^vector (v)/dt so we need to differentiate the given position vector with respect to time. V^vector (t) = d/dt [(4t + 6t^2) i^^+ (-4 + 1.5 t^2) J^^] = [4 d/dt (t) + 6 d (t^2) /dt] i^^+ [d/dt (-4) + 1.5 d (t^2) /dt] J^^= (4 times 1 + 6 times 2t) i^^+ (0 + 1.5 times 2t) J^^V^vector (t) = (12t + 4) i^^+ (3t) J^^m/s Acceleration of the as a function

Answer 4

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Answer 5

The position of an object is r (t) = (4t + 6t^2) I + (1.5t^2 - 4) j The velocity of the object is v (t) = dx (t)/dt = d/dt ((4t + 6t^2) i + (1.5t^2 - 4)j) = (4 + 12t) I + 3.0 tj The acceleration of the object is a (t) = d/dt (t) = d/dt ((4 + 12t) I + 3.0 tj) = 12 i + 3.0 j The magnitude of the instantaneous velocity at t = 3.0 s is v (t) = (4 + 12t) I + 3.0 tj = (4 + 12 (3.0)) I + 3.0 (3.0) j = 40i + 9j The direction of an acceleration is phi = tan^-1 (3/12) = 14.036 degree