Question

The displacement of a particle moving along x- axs is given by x=15t + 3t² calculate
i.)average velocity b/w t=2s & t=3s
ii.) instantaneous acceleration


gyus answer it fast plsss​

Answer 1

Answer:

see above

instantaneous acc is at instant of time and the instant of time is not given in ques.

hope so helpful

Answer 2

Answer:

• The average velocity (Δ v) is 30 m/s.
• The Instantaneous acceleration (a) is 3 m/s²

Given:

1. Given equation, x = 15 t + 3 t²

Explanation:

$\rule{300}{1.5}$

Average velocity = ?

For this we need to find first Displacement at t = 2 second and t=3 second respectively.

Displacement at t = 2 sec

$\dashrightarrow\sf x=15\;t + 3\;t^{\;2}\\\\\\\dashrightarrow\sf x_{\;1} = 15\times (2) +3\times (2)^{\;2}\ \ \because \big\lgroup t_{\;1}=2\big\rgroup \\\\\\\dashrightarrow\sf x_{\;1} =30+3\times4\\\\\\\dashrightarrow\sf x_{\;1}=30+12\\\\\\\dashrightarrow \underline{\sf x_{\;1}=42\;m}$

Displacement at t = 3 sec

$\dashrightarrow\sf x=15\;t + 3\;t^{\;2}\\\\\\\dashrightarrow\sf x_{\;2} = 15\times (3) +3\times (3)^{\;2}\ \ \because \big\lgroup t_{\;2}=3\big\rgroup\\\\\\\dashrightarrow\sf x_{\;2} =45+3\times9\\\\\\\dashrightarrow\sf x_{\;2}=45+27\\\\\\\dashrightarrow \underline{\sf x_{\;2}=72\;m}$

Now, finding the average velocity,

$\dashrightarrow\sf Average\;Velocity=\dfrac{Final\;Disp.-Initial\; Disp.}{Total\;time}$

Substituting the values,

$\dashrightarrow\sf \Delta\;v=\dfrac{x_{\;2}-x_{\;1}}{t_{2}-t_{\;1}}\\\\\\\dashrightarrow\sf \Delta\; v =\dfrac{72-42}{3-2}\\\\\\\dashrightarrow\sf \Delta\; v =\dfrac{30}{1}\\\\\\\dashrightarrow \large{\underline{\boxed{\red{\sf v=30\;m/s}}}}$

∴ The average velocity (Δ v) is 30 m/s.

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Instantaneous Acceleration = ?

Differentiating two times the given equation we will get Inst. acceleration.

Differentiating,

$\displaystyle \dashrightarrow\sf x=15\;t + 3\;t^{\;2}\\\\\\\dashrightarrow\sf \dfrac{d\;x}{d\;t}=\dfrac{d\;(15\;t+3\;t^{\;2})}{d\;t}\\\\\\\dashrightarrow\sf v=\dfrac{d\;(15\;t+3\;t^{\;2})}{d\;t}\\\\\\\dashrightarrow\sf v=15+3\;t^{\;(2-1)}\\\\\\\dashrightarrow\sf v=15+3\;t$

Again differentiating,to get acceleration,

$\displaystyle \dashrightarrow\sf v=15 + 3\;t\\\\\\\dashrightarrow\sf \dfrac{d\;v}{d\;t}=\dfrac{d\;(15+3\;t)}{d\;t}\\\\\\\dashrightarrow\sf a=\dfrac{d\;(15+3\;t)}{d\;t}\\\\\\\dashrightarrow\sf a=0+3\;t^{\;(1-1)}\\\\\\\dashrightarrow\sf a=3\;t^{\;0}\\\\\\\dashrightarrow \large{\underline{\boxed{\red{\sf a=3\;m/s^{\;2}}}}}$

∴ The Instantaneous acceleration (a) is 3 m/s².

$\rule{300}{1.5}$