Question

A particle is moving along a straight line with an initial velocity of 3 m/s when it is subjected to a deceleration of a = - 1.1 v^1/2 m/s^2​​ .A. Determine how far it travels before it stops.B. How much time does it take?

Answer 1

Step-by-step explanation:

Given that

initial velocity ,
`v= 3 m/s`

`a=-1.1v^{(1)/(2)}`

We know that

`a=v(dv)/(dx)`

Lets take x is the distance before coming to the rest.

The final speed of the particle = 0 m/s

`v(dv)/(dx)=-1.1v^{(1)/(2)}`

`(dv)/(dx)=-1.1v^{-(1)/(2)}`

`v^{(1)/(2)}{dv}=-1.1dx`

`\int_(3)^(0)v^{(1)/(2)}{dv}=-\int_(0)^(x)1.1dx`

`\left [v^{(3)/(2)}* (2)/(3)\right]_3^0=-1.1x`

`3^{(3)/(2)}* (2)/(3)=1.1x`

`x=(3.46)/(1.1)\ m\\x=3.14\ m`

(b)time taken by it

`a=\frac{\mathrm{d} v}{\mathrm{d} t}=-1.1√(v)`

`\int_(3)^(0)(dv)/(√(v))=-1.1\int_(0)^(t)dt`

`\int_(0)^(3)(dv)/(√(v))=1.1\int_(0)^(t)dt`

`2* 3√(3)=1.1t`

`t=9.44\ s`