Question

Draw the velocity _ time graph for the following situation 1. When the body is moving with variable velocity but with uniform accleration

Answer 1

Answer:

Given:

Body is moving with uniform Acceleration ; Hence has variable velocity

To find:

Velocity - time graph for the same body

Calculation:

As per the question, the acceleration is constant. Mathematically, we can say :

$acceleration = k \: \: m {s}^{ - 2}$

Since we are trying to find out the instantaneous velocity, we can say :

$\implies \: \dfrac{dv}{dt} = k$

$\implies \: dv = k \: dt$

Integrating on both sides :

$\displaystyle \implies \: \int dv = \int \: k \: dt$

Putting the limits :

$\displaystyle \implies \: \int_{0}^{v} dv = \int_{0}^{t} \: k \: dt$

$\implies \: v - 0 = k \: \{t - 0 \}$

$\implies \: v = kt$

So it's a linear relationship passing through origin and zero intercept . Considering positive acceleration, the slope will be positive.

Answer 2

SolutioN :

• Body is moving with variable velocity but with uniform acceleration

Let the acceleration be x m/s²

$\dashrightarrow \tt{a \: = \: x} \\ \\ \dashrightarrow \tt{\dfrac{dv}{dt} \: = \: x} \\ \\ \dashrightarrow \tt{dv \: = \: x dt} \\ \\ \dashrightarrow \tt{\int dv \: = \: \int xdt} \\ \\ \dashrightarrow \tt{\int_0 ^v dv \: = \: x \int_0 ^t dt} \\ \\ \dashrightarrow \tt{[v \: - \: 0] \: = \: x[t \: - \: 0]} \\ \\ \dashrightarrow \tt{v \: = \: xt} \\ \\ \dashrightarrow \tt{v \: \propto \: t} \\ \\ \boxed{\sf{Where, \: x \: is \: constant}}$

• Acceleration is defined as rate of change of velocity.

• It's SI unit is m/s².

• Acceleration is a vector quantity.

• It can be 0, positive or may be negative as well