Question

Heyy guys...
if position Vs time equation of particle is y=6t^2-2t+10,where y is in metre and t in sec. find:
a)Initial position of particle
b) velocity after 3 sec
c) initial velocity of particle
d) acceleration after 2 sec
e) initial acceleration of particle
f) velocity of particle when acceleration is zero
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Answer 1

Answer:

a) 10 m

b) 34 m/s

c) -2 m/s(minus implies backwards or opposite direction)

d) 12 m/s²

e) 12 m/s²

f) Not defined or does not exist or hypothetical

Explanation:

y = 6t²-2t+10.............(1)

Differentiate with respect to x,

dy/dx= 12t-2(velocity, v).....(2)       {Math help: d/dx (x^n) = n x^n-1}

Again differentiate,

d²y/dx² = 12 (Acceleration, a).........(3)

a) For initial position t=0, because that's when the motion is started.

From(1)

∴y = 10 m

i.e initial position of particle is 10 m

b)Given,t = 3 sec

From(2)

v = 12(3) - 2

v = 34 m/s

therefore velocity after 3 seconds is 34 m/s

c) Initially, t=0

From(2)

v = 12(0) - 2

v = -2 m/s

i.e initial velocity is -2 m/s(minus implies backwards or opposite direction)

d & e) Acceleration (3) has no variable dependence i.e it does not change with time.

Therefore acceleration of that particle at t=0 or t=2 or any other time remains constant i.e 12 m/s².

f) It's a hypothetical question. Because acceleration can never be zero in this case.