TS EAMCET 5-Aug-2021 Shift 2 Question Paper

Given, Equation of motion of particle, y=4x2 . . . (i)
Velocity of particle, v=2

∧

i

+2

∧

j

∴‌‌vx=2m∕s and vy=2m∕s
and acceleration, a=a

∧

j

⇒‌‌ay=am∕s2
As we know that,
Velocity, V=‌

dy

dt

or ‌

dx

dt

along Y-axis and X-axis respectively.
On differentiating Eq. (i) both sides with respect to t, we get,
‌‌

dy

dt

‌‌=4×2x‌

dx

dt

⇒‌vy‌‌=8xvx.(ii)
⇒‌2‌‌=8x×2
⇒‌x‌‌=‌

1

8

m
On differentiating Eq. (ii) with respect to time t, we get
‌‌

dvy

dt

‌‌=8vx‌

dx

dt

+8x‌

dvx

dt

⇒‌‌ay‌‌=8vx⋅vx+8xax=8vx2+8xax
⇒‌a‌‌=8×22+8×‌

1

8

×0
⇒‌‌a‌‌=32m∕s2