Question

The velocity of a body at a given instant is given by V= 3i+(4-2t)k
(A)-what is the magnitude and direction ofvthe initial velocity of the body
(B)- at what instant the body will hit x axis again
(C)-what is the shape of trajectory and why?
(D)- what is maximum distance covered in y axis

Answer 1

Answer on Question #57430-Physics-Mechanics

The velocity of a body at a given instant is given by $V = 3i + (4 - 2t)k$

(A)-what is the magnitude and direction of the initial velocity of the body

(B)-at what instant the body will hit x axis again

(C)-what is the shape of trajectory and why?

(D)-what is maximum distance covered in y axis

Solution

(A) Initial velocity is when $t = 0$.

Substituting this into the equation above:

$$\ddot{V}(0) = 3\ddot{i} + (4 - 2 \cdot 0)\ddot{k} = 3\ddot{i} + 4\ddot{k}.$$

The magnitude of the initial velocity of the body is

$$V = \sqrt{3^2 + 4^2} = 5.$$

The direction of the initial velocity of the body is

$$\theta = \tan^{-1} \frac{4}{3} = 53{}^\circ \text{ with the } x \text{ axis}.$$

(B) The body will hit x axis again when $y = 0$.

$$y(t) = \int_0^t (4 - 2s)\mathrm{d}s = 4t - 2\frac{t^2}{2} = 4t - t^2 = t(4 - t).$$

The time will be

$$t = 4.$$

(C) $z = 4t - t^2$

$$x = 3t \rightarrow t = \frac{x}{3}.$$ $$y = 4\left(\frac{x}{3}\right) - \left(\frac{x}{3}\right)^2 = \frac{4}{3}x - \frac{1}{9}x^2.$$

This is equation of parabola.

(D) Maximum distance covered in y axis will be at

$$\frac{dy}{dt} = 4 - 2t = 0 \rightarrow t = 2.$$ $$y_{\max} = 2(4 - 2) = 4.$$

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