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.
Answers
Answe ◘
Given
Velocity of a car = 2\sqrt{I}2
I
where I = distance from a fixed point.
To Find
Acceleration of the car when I = 8 m.
Concept
\begin{gathered}v = \frac{dx}{dt} \\ \\ \\ a = \frac{dv}{dt} \\ \\ \\ a = v \frac{dv}{dx}\end{gathered}
v=
dt
dx
a=
dt
dv
a=v
dx
dv
These type of questions can be solved with the help of differentiation and integration as they are not linear motion and hence,equations of motion are not applicable.
The above listed formulas uses instantaneous concepts.
V = 2\sqrt{I}2
I
implies velocity of the car is a function of the distance from fixed point.As distance from that fixed point changes,velocity changes.
Calculations
Assume I to be x. So,we have to calculate acceleration at x = 8 m.
Given,V = 2\sqrt{x}2
x
Now, a = v\dfrac{dv}{dx}v
dx
dv
\begin{gathered}=\:v \dfrac{d(2\sqrt{x})}{dx} \\ \\ \\ = \:v \dfrac{2}{2\sqrt{x}} \\ \\ \\ = 2 \sqrt{x} \times \dfrac{1}{\sqrt{x}}\\ \\ \\ =\: 2\end{gathered}
=v
dx
d(2
x
)
=v
2
x
2
=2
x
×
x
1
=2
Hence,acceleration of the car will be 2 m/s² at that instant.
Explanation:
hope it's helpful to you