Question

A body moves in a straight line under the retardation a which is given by a = bv^2 . if the initial velocity is 2m/s , the distance covered in t= 2 seconds is

Answer 1

Answer:

1/b ln (1+4b)

Explanation:

use integration formulas

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Answer 2

Answer:

The distance covered in t= 2 seconds will be $\frac{1}{b}$ ln(1 + 4b).

Explanation:

Given that straight line under the retardation is a = -bv².

$\frac{dv}{dt}$ = -bv²

$\frac{dv}{v^{2} }$ = -bdt

∫$\frac{dv}{v^{2} }$ = -∫bdt

- $\frac{1}{v}$ = -bt + c

At t = 0, v = 2 m/s

-$\frac{1}{2}$ = c

-$\frac{1}{v}$ = -bt - $\frac{1}{2}$

$\frac{1}{v}$ = $\frac{1+2bt}{2}$

v = $\frac{2}{1+2bt}$

$\frac{ds}{dt}$ = $\frac{2}{1+2bt}$ dt

∫ds = ∫$\frac{2}{1+2bt}$ dt

s = $\frac{2ln(1+2bt)}{2b}$

s  = $\frac{1}{b}$ln(1 + 2bt)

When t = 2s

So, s(2) = $\frac{1}{b}$ln(1 + 4b)

The above solution is based on the Motion in a Straight Line analysis.

What is Motion in a Straight Line?

An object is said to be in motion if its position in relation to its surroundings changes over time. It is a shift in an object's position over time. Linear motion is the only type of motion that exists. Any object that moves from one point to another in relation to the observer is said to be in motion. Motion in a Straight Path-Rectilinear motion is the term used to describe an object's motion when it moves in a straight line. For instance, the movement of an automobile on a roadway.

Thus, one-dimensional motion is defined as the movement of a particle along a path or in a straight line.