Question

A particle is moving with initial velocity 1 m/s and acceleration a = (4t + 3) m/s?
Find velocity of particle at t= 2 sec.​

Answer 1

Given:

Initial velocity of the particle= 1 m/s

Acceleration applied a = (4t + 3) m/s²

To find:

The velocity of the particle at t = 2 sec

Solution:

Since acceleration is not constant we need to proceed in the following manner:

We know that a = dv/dt

Where v is the velocity of the particle,

a is the acceleration,

t is the time taken.

dv = a dt

On integrating we get the following equation:

$\int\limits^v_1 {} \, dv=$$\int\limits^2_0 {a} \, dt$

v is the final velocity of the particle.

$(v-1)=4\int\limits^2_0 {t} \, dt +3\int\limits^2_0 {} \, dt$

(v - 1) = 4 [t²/ 2]₀² + 3 [t]₀²

(v - 1) = 4*(2-0) + 3*(2-0)

(v - 1) = 8 + 6

v = 14 + 1

v = 15 m/s

Therefore, the velocity at t = 2 sec will be 15 m/s.

Answer 2

SOLUTION

GIVEN

A particle is moving with initial velocity 1 m/s and. acceleration a = (4t + 3) m/s²

TO DETERMINE

The velocity of particle at t = 2 sec.

EVALUATION

Here it is given that acceleration a = (4t + 3) m/s²

If v m/s is the velocity of the particle then

$\displaystyle \sf{a = \frac{dv}{dt} }$

$\displaystyle \sf{ \implies \frac{dv}{dt} = 4t + 3 }$

On integration we

$\displaystyle \sf{v = \int\limits_{}^{} (4t + 3) \, dt}$

$\displaystyle \sf{ \implies \: v = 2 {t}^{2} + 3t + c} \: \: \: ....(1)$

Where c is a integration constant

Here it is also stated that the particle is moving with initial velocity 1 m/s

So at t = 0 we have v = 1

Equation (1) gives

$\displaystyle \sf{ 1 = 2 {(0)}^{2} + 3 \times 0 + c} \: \: \:$

$\displaystyle \sf{ \implies \: c = 1} \: \: \:$

So Equation (1) becomes

$\displaystyle \sf{ \: v = 2 {t}^{2} + 3t + 1} \: \: \:...(2)$

So the velocity of the particle at t = 2 is

$\displaystyle \sf{ \: v \big|_{t = 2} = 2 \times {(2)}^{2} +( 3 \times 2) + 1} \: \: \:$

$\displaystyle \sf{ \implies \: v \big|_{t = 2} = 8 + 6 + 1}$

$\displaystyle \sf{ \implies \: v \big|_{t = 2} = 15}$

FINAL ANSWER

Hence the velocity of the particle at t = 2 sec is 15 m/s

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