A particle moves along the curve `y^(2) = 2x` where `x = (t^(2))/(2)` & `y gt 0`. What is the acceleration of the particle at `t = 2s` ?

Question

A particle moves along the curve `y^(2) = 2x` where `x = (t^(2))/(2)` & `y gt 0`. What is the acceleration of the particle at `t = 2s` ?

A particle moves along the curve `y^(2) = 2x` where `x = (t^(2))/(2)` & `y gt 0`. What is the acceleration of the particle at `t = 2s` ?
A. `hat(i)`
B. `hat(j)`
C. `hat(i) - hat(j)`
D. `2 hat(i)`

Monjur Alahi

6 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - A
`x = (t^(2))/(2)`
`v_(x) = (2t)/(2) rArr a_(x) = hat(i)`
`y^(2) = 2 xx (t^(2))/(2)`
`rArr y = t rArr a_(y) = 0`
`vec(a) = vec(a)_(x) + vec(a)_(y) = hat(i)`

6 views Answered 2023-02-05 15:29