Question

if the displacement of a particle after t seconds is given by s=2t^3-3t^2+2,find position,velocity and acceleration at the end of 2 seconds.

Answer 1

s=2t^3-3t^2+2
s=16-12+2=6

v=ds/dt= 6t^2-6t
v=24-12=12

a=dv/dt = 12t-6
a=24-6=18

Answer 2

$\huge{\bigstar}{ \green{ \mathfrak{ \underline{ \underline{Question}}}}}{\bigstar}$

□☆□☆□☆□☆□☆□☆□☆□☆□☆□☆□☆

if the displacement of a particle after t seconds is given by s = 2t³- 3t² + 2, find position,velocity and acceleration at the end of 2 seconds.

□☆□☆□☆□☆□☆□☆□☆□☆□☆□☆□☆

──────────────────────

$\huge{\bigstar}{ \green{ \mathfrak{ \underline{ \underline{Answer}}}}}{\bigstar}$

$\boxed{\red{\bold{Given:}}}$

$\red{\implies} s = 2t^3 - 3t^2 + 2$

$\boxed{\red{\bold{The \ position \ after \ 2 \ s:}}}$

$\purple{\implies}s = 2(2)^3 - 3(2)^2 +2$

$\purple{\implies}s = 2(8) - 3(4) +2$

$\purple{\implies}s= 16 - 12 +2$

$\purple{\implies}s= 6$

$\boxed{\red{\bold{Velocity \ after \ 2 \ s}}}$

$\blue{\implies}v = \displaystyle\frac{ds}{dt}$

$\blue{\implies}v = \displaystyle\frac{d}{dt} 2t^3 - 3t^2 + 2$

$\blue{\implies}v = 6t^2 - 6t$

$\blue{\implies}v = 6(2)^2 - 6(2)$

$\blue{\implies}v = 24 - 12$

$\blue{\implies}v = 12$

$\boxed{\red{\bold{Acceleration \ after \ 2 \ s}}}$

$\green{\implies}a = \displaystyle\frac{dv}{dt}$

$\green{\implies}a = \displaystyle\frac{d}{dt} 6t^2 - 6t$

$\green{\implies}a = 12t - 6$

$\green{\implies}a = 12(2) - 6$

$\green{\implies}a = 24 - 6$

$\green{\implies}a = 18$

──────────────────────