Starting at time t = 0 from the origin with speed 1 ms^-1, a particle follows a two – dimensional trajectory in the x – y plane so that its coordinates are related by the equation $y = \frac{x^{2}}{2}$ . The x and y components of its acceleration are denoted by a_x and a_y, respectively. Then

see full answer

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

a_x = 0 implies a_y = 1ms^-2 at all times

a_x = 1ms^-2 implies that when the particle is at the origin, a_y = 1ms^-2

at t = 0, the particle’s velocity points in the x – direction

a_x = 0 implies that at t = 1 s, the angle between the particle’s velocity and the x axis is 45⁰

answer is A, B, C, D.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$y = \frac{x^{2}}{2}$

$v_{y} {= x v}_{x}$

$a_{y} {=xa}_{x} {+v}_{x}^{2}$

$x = 0 \Rightarrow a_{y} {=v}_{x}^{2} {= 1m/s}^{2}$

For any value of a_x

$a_{x} = 0 \Rightarrow a_{y} {= v}_{x}^{2} =1$

$tan θ= \frac{V_{y}}{V_{x}} = x =1$

x = v_x t = 1

Watch 3-min video & get full concept clarity