The horizontal and vertical displacements of a particle moving along a curved line are given by x=5t and=22+ .Time after which its velocity vector makes an angle of 45° with the horizontal is

vx=dxdt=5 vy=dydt=(4t+1) at 45o vx=vy t=1 s

The horizontal and vertical displacements of a particle moving along a curved line are given by x=5t and=22+ .Time after which its velocity vector makes an angle of 45° with the horizontal is

vx=dxdt=5 vy=dydt=(4t+1) at 45o vx=vy t=1 s

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The horizontal and vertical displacements of a particle moving along a curved line are given by $x = 5 t$ and $𝑦 = 2 𝑡^{2} + 𝑡$ .Time after which its velocity vector makes an angle of 45° with the horizontal is

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1 s

0.5 s

1.5 s

2 s

answer is B.

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Detailed Solution

$v_{x} = \frac{d x}{d t} = 5 v_{y} = \frac{d y}{d t} = (4 t + 1) a t 45^{o} v_{x} = v_{y} t = 1 s$

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