Question

If the velocity at x=0m is 0.8m/s, find the velocity at x=1.4m.​

Answer 1

Since in the graph, acceleration is variable with 'x' , we can't use equation of motion.

so use generally equation of acceleration.

$\rightarrow a = \dfrac{dv}{dt}$

• Expand it more:

$\implies \rm a= \dfrac{dv}{dt} =\dfrac{dv\times dx}{dx\times dt}$

$\implies \bf a= v \dfrac{dv}{dx}$

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• In the graph we can use the equation of y=mx+c
• slope will be same for distance 0.8/1.4 m, so just take that 0.4 to 0.8 for calculating slope , as acceleration is same for both.

$\longrightarrow \rm y = mx+ c$

$\longrightarrow \rm a = \dfrac{dy}{dx}x$

$\longrightarrow \rm a = \dfrac{0.4-0.2}{0.8-0.4}x$

$\longrightarrow \rm a = \dfrac{0.2}{0.4}x$

$\longrightarrow \rm a = \dfrac{1}{2}x$

$\longrightarrow \rm v\dfrac{dv}{dx} = \dfrac{1}{2}x$

Integrate it by putting limits:

$\longrightarrow vdv = \dfrac{1}{2}xdx$

$\longrightarrow \displaystyle \rm 2\int vdv = \int xdx$

$\longrightarrow \displaystyle \rm 2\int_{0.8} ^{v} \dfrac{v^2}{2} = \int_{0}^{1.4} \dfrac{x^2}{2}$

$\longrightarrow \rm 2(v^2 -0.8^2) = 1.4^2-0^2$

$\longrightarrow \rm 2(v^2 -0.64) = 2.96$

$\longrightarrow \rm v^2 -0.64 = 1.48$

$\longrightarrow \rm v^2 = 2.12$

$\longrightarrow \rm v = \sqrt{2.12}$

$\longrightarrow \rm v \approx 1.46\: ms^{-1}$

☆so velocity at x=1.4 is 1.46m/s.