Question

The velocity of a body at a given instant t is given by v = 3i + (4 – 2t)j (See lesson 4)
(i) What is the magnitude and direction of the initial velocity of the body?
(ii) At what instant will the body hit the x-axis again?
(iii) What is the shape of the trajectory? why?
(iv) What is the maximum distance moved by the body along the y-axis?

Answer 1

Answer on Question #57463-Physics – Mechanics | Relativity

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

(i) What is the magnitude and direction of the initial velocity of the body?

(ii) At what instant will the body hit the $x$-axis again?

(iii) What is the shape of the trajectory? why?

(iv) What is the maximum distance moved by the body along the $y$-axis?

Solution

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

Substituting this into the equation above:

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

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 $y = 0$.

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

The time will be

$$t = 4.$$

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

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

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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