Question

Velocity of a car depends on its distance l from a fixed pole on a straight road as v = 2√l, where l is in metres and v in m/s. Find acceleration (in m/s²) when l = 8m.​​

Answer 1

Answer:

◘ Given ◘

Velocity of a car depends on its distance l from a fixed pole on a straight road as v = 2√l, where l is in metres and v in m/s.

$\bullet\ \sf{|\overrightarrow{v}|=2\sqrt{l}\ m/s}$

◘ To Find ◘

The acceleration (in m/s²) when l = 8m.

◘ Solution ◘

$\sf{\overrightarrow{a}=\dfrac{d \overrightarrow{v}}{dt}} </p><p></p><p>$

► Car is moving along a straight line without any change in its direction. So, change in magnitude of velocity due to tangential acceleration :

$\begin{gathered}\sf{\overrightarrow{a}_t=\dfrac{d}{dt} |\overrightarrow{v}|}\\\\\sf{\longrightarrow \overrightarrow{a}_t=|\overrightarrow{v}| \dfrac{d|\overrightarrow{v}|}{dx} }\end{gathered} </p><p></p><p> </p><p>$

_________________

From above,

$\begin{gathered}\sf{\dfrac{d|\overrightarrow{v}|}{dx}=\dfrac{d |\overrightarrow{v}|}{dl}=\dfrac{dv}{dl}}\\\\\sf{=\dfrac{d}{dl}.\sqrt{2}l}\\\\\sf{=\dfrac{1}{\sqrt{l}}}\end{gathered}$

_________________

$\begin{gathered}\sf{a_t=2 \sqrt{l} \times \dfrac{1}{\sqrt{l}}}\\\\\boxed{\sf{\longrightarrow a_t=2\ m/s^2}}\end{gathered} </p><p>$

Car's motion is uniformly accelerated. So, acceleration at l = 8 m is 2 m/s².

Answer 2

$\frak \purple{ Refer \: to \: the \: attachment}$