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Zepler [3.9K]
3 years ago
9

A particle moving along a straight line is subjected to a deceleration ???? = (−2???? 3 ) m/???? 2 , where v is in m/s. If it ha

s a velocity v = 10 m/s and a position s = 15 m when t = 0, determine its velocity and position when t = 25 s.
Physics
1 answer:
dlinn [17]3 years ago
3 0

Answer:

(a). The velocity is 0.099 m/s.

(b). The position is 19.75 m.

Explanation:

Given that,

The deceleration is

a=(-2v^3)\ m/s^2

We need to calculate the velocity at t = 25 s

The acceleration is the first derivative of velocity of the particle.

a=\dfrac{dv}{dt}

\dfrac{dv}{dt}=-2v^3

\dfrac{dv}{-v^3}=2dt

On integrating

int{\dfrac{dv}{-v^3}}=\int{2dt}

\dfrac{1}{2v^2}=2t+C

v^2=\dfrac{1}{4t+2C}....(I)

At t = 0, v = 10 m/s

10^2=\dfrac{1}{4\times0+2C}

C=\dfrac{1}{200}

Put the value of C in equation (I)

v^2=\dfrac{1}{4\times25+2\times\dfrac{1}{200}}

v=\sqrt{\dfrac{1}{4\times25+2\times\dfrac{1}{200}}}

v=0.099\ m/s

The velocity is 0.099 m/s.

(b). We need to calculate the position at t = 25 sec

The velocity is the first derivative of position of the particle.

\dfrac{ds}{dt}=v

On integrating

\int{ds}=\int(\sqrt{\dfrac{200}{800t+1}})dt

s=\dfrac{\sqrt{200}\times2\sqrt{800t+1}}{800}+C'

At t = 0, s = 15 m

15=\dfrac{200}{800}+C'

C'=15-\dfrac{200}{800}

C'=14.75

Put the value in the equation

s=\dfrac{\sqrt{200}\times2\sqrt{800\times25+1}}{800}+14.75

s=19.75\ m

The position is 19.75 m.

Hence, (a). The velocity is 0.099 m/s.

(b). The position is 19.75 m.

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