Question

The horizontal and vertical displacements of a particle moving along a curved line are given by $x = 5t$ and $y = 2t^2 + t$ (here $x$ & $y$ are in m and $t$ is in s). Time after which its velocity vector makes an angle of $45^\circ$ with the horizontal is

• A. 0.5 s
• B. 1 s
• C. 2 s
• D. 1.5 s

Answer 1

Answer: (B)

1 s

Answer 2

To find the time when the velocity vector makes an angle of $45^\circ$ with the horizontal, we first need to determine the horizontal and vertical components of the velocity.

Given the horizontal displacement: $x = 5t$ m

The horizontal component of velocity ($v_x$) is the time derivative of $x$: $v_x = \frac{dx}{dt} = \frac{d}{dt}(5t) = 5 \, \text{m/s}$

Given the vertical displacement: $y = 2t^2 + t$ m

The vertical component of velocity ($v_y$) is the time derivative of $y$: $v_y = \frac{dy}{dt} = \frac{d}{dt}(2t^2 + t) = 4t + 1 \, \text{m/s}$

The velocity vector $\vec{v}$ can be written as $\vec{v} = v_x \hat{i} + v_y \hat{j}$. The angle $\theta$ that the velocity vector makes with the horizontal is given by: $\tan \theta = \frac{v_y}{v_x}$

We are given that the velocity vector makes an angle of $45^\circ$ with the horizontal, so $\theta = 45^\circ$. We know that $\tan 45^\circ = 1$.

Therefore, we can set up the equation: $1 = \frac{4t + 1}{5}$

Now, solve for $t$: $5 = 4t + 1$ $5 - 1 = 4t$ $4 = 4t$ $t = \frac{4}{4}$ $t = 1 \, \text{s}$

Thus, the time after which its velocity vector makes an angle of $45^\circ$ with the horizontal is 1 second.