Question

A body moves with its position varying as (mathbfR = (4t² - 16t)mathbfi + 3t²mathbfj) meters. What is the y-component of its velocity when the x-component is zero? Find also the position vector of the body at this instant.

a) (V_y = 0), (mathbfR = -64mathbfi + 48mathbfj)
b) (V_y = -16), (mathbfR = -32mathbfi + 12mathbfj)
c) (V_y = -48), (mathbfR = 0mathbfi + 0mathbfj)
d) (V_y = -32), (mathbfR = -64mathbfi + 0mathbfj)

Answer 1

The y-component of the velocity when the x-component is zero is 24 m/s. The position vector of the body at this instant is -64i + 48j.

Step-by-step explanation:

To find the y-component of the velocity when the x-component is zero, we need to find the derivative of the position vector with respect to time. Taking the derivative, we get v(t) = 8t - 16i + 6tj. At the instant when the x-component is zero (i.e., when 4t² - 16t = 0), we can solve for t to find that t = 0 or t = 4. Plugging in t = 4 into the velocity vector, we get v(4) = 32i + 24j. Therefore, the y-component of the velocity when the x-component is zero is 24 m/s.

To find the position vector at this instant, we can integrate the velocity vector with respect to time. Integrating v(t) = 8t - 16i + 6tj, we get r(t) = 4t² - 16t + C1i + 3t² + C2j, where C1 and C2 are constants. Since the initial position is not given, we can find the position vector at t = 4 by determining the values of C1 and C2. Plugging in t = 4 and the given position vector (4t² - 16t)i + 3t²j, we can solve for C1 and C2. The resulting position vector is -64i + 48j.