Question

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

Answer 1

1) Initial velocity = v(t = 0) = 3i + 4j.
Magnitude =
$\sqrt{ {3}^{2} + {4}^{2} } = 5$
Direction = along the unit vector (3/5)i + (4/5)j.

Since nothing has been said, let's assume that the body is starting from the origin.

2) The body will hit the x-axis when y = 0.
v along y = dy/dt
4 - 2t = dy/dt
dy = (4 - 2t) dt
Integrating both sides (y = 0 to y, t = 0 to t)
y = 4t - t^2
As said, when y = 0, the body will hit the x-axis again.
So, 4t - t^2 = 0 or t = 0 or 4.
After t = 0, the body will again hit the x-axis when t = 4.

3) v along x = dx/dt
Similar to (2), we again have x = 3t.
We also have y = 4t - t^2.
Eliminating t, the equation of the trajectory of the body is y = 4(x/3) - (x/3)^2, which is that of a downward opening parabola.

4) y is maximum when dy/dt = 0.
That is, when v along y is 0.
Or, when 4 - 2t = 0.
Or, when t = 2.
Hence, distance covered along y axis till t = 2 is 4(2) - 2^2 = 4 units, which the maximum distance.