Question

The acceleration of a particle in m/s2 is given by a = 4t 2+2t+2, where time t is in second. If the particle starts with a velocity v = 2 m/s at t=0, find the velocity at t = 5 s.​

Answer 1

Question

The acceleration of a particle in m/s2 is given by a = 4t 2+2t+2, where time t is in second. If the particle starts with a velocity v = 2 m/s at t=0, find the velocity at t = 5 s.

$\begin{gathered} \\ {\underline{\rule{200pt}{3pt}}} \end{gathered}$

Answer

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Information provided with us -

$\sf\mapsto{\underline{\boxed{\green{{Given ,\: a= \bigg(\dfrac{d \: v}{dt }\bigg) = {4 \: t}^{2} + 2 \: t + 2 }}}}}$

${\large{\underline{\pmb{\frak{OR}}}}}$

$\sf\mapsto{\underline{\boxed{\red{{ \: d \: v = {(4 \: t}^{2} + 2 \: t + 2) \: d \: t }}}}}$

$\begin{gathered} \\ {\underline{\rule{200pt}{3pt}}} \end{gathered}$

What we have to find out :

• The velocity at t = 5 s.

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• When the time changes from 0 to 5 s the velocity changes from 2 ms-¹ to v

• By Integrating the above relation within the condition of motion we will get required answer

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${\large{\underline{\pmb{\frak{Calculation}}}}}$

$\rm\implies \:Given : 4 \: {t}^{2} + 2 \: t + 2$

$\displaystyle\implies \: \rm \because\: \: a = \dfrac{d \: v}{d \: t}$

$\displaystyle\implies \: \rm \dfrac{d \: v}{d \: t} = 4 \: {t}^{2} + 2 \: t + 2$

$\displaystyle\implies \: \rm d \:v= (4 \: {t}^{2} + 2 \: t + 2) \: d \: t$

$\displaystyle\implies \rm \int\limits_{u = 2}^{v = v}dv = \int\limits_{t = 0}^{t = 5}( \: 4 \: {t}^{2} + 2 \: t + 2) \: d \: t$

$\rm{\displaystyle\implies \left \bigg\{v\bigg\}\right_{2}^{v} = \bigg\{\dfrac{4 \: {t}^{3} }{3} + \dfrac{2 \: {t}^{2} }{2} + 2 \: t \bigg\}\right_{0}^{5}}$

• Here 2 and 2 get cancelled out

$\rm\displaystyle\implies \: (v - 2 )= \bigg\{\dfrac{4 \: {t}^{3} }{3} + {t}^{2} + 2 \: t \bigg\}\right_{0}^{5}$

$\rm\displaystyle\implies \: (v - 2 )= \dfrac{4}{3} \times ( {5)}^{2} + ( {5)}^{2} + 2 \times 5$

$\rm\displaystyle\implies \: (v - 2 )= (\dfrac{4}{3} \times 125) + (25) + 10$

$\rm\displaystyle\implies \: (v - 2 )= \dfrac{500}{3} + 35$

$\rm\displaystyle\implies \: (v - 2 )= (166.67 + 35) = 201.67$

$\rm\displaystyle\implies \: v= (2 + 201.67 ) = (203.67)$

$\displaystyle\implies \bf v = 203.67 \: m /s$

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Required Answer

$\displaystyle\implies \underline{ \boxed{ \red{ \bf v = 203.67 \: m/s}}}$

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Answer 2

Explanation:

Another kinematics question! Well,tbh I won't even like to put this question in kinematics, instead this can be said to be a question basically from basic mathematics,why? Because here I suppose the only tricky part is that students can't figure out what actually they are required to do, differentiation or integration? At least we are sure about the fact that we have to use either of them because the acceleration is dependent on the time factor,which means it is going change with time and that's where we actually see calculus coming into picture.

So, the basic thing you should know :

If you have been provided with velocity in the question and you are asked to calculate acceleration, you should differentiate velocity wrt t to obtain acceleration, because we know :

a = dv/dt

And if you been provided with acceleration in the question (you can consider this question as an example of this kind of case) and asked to calculate velocity,we integrate, because :

a = dv/dt

adt = dv

∫dv = ∫adt

v = ∫adt

So I guess we have discussed already sufficient about the procedure to be followed, let's now solve the question.

Given that a = 4t² + 2t + 2

Clearly the acceleration depends on time factor,so calculus comes into the picture.

And the velocity of the particle at t = 0 m/s is given to be v = 2 m/s,we are required to find the velocity at t = 5 seconds.

Using the above explained concept,we can write :

a = dv/dt

4t² + 2t + 2 = dv/dt

(4t² + 2t + 2)dt = dv

∫(4t² + 2t + 2)dt = ∫dv (Upper limit = v, lower = 2)

(t's upper limit = 5, lower limit = 0)

V = ∫4t²dt + ∫2tdt + ∫2dt

(Using : ∫xⁿ = xⁿ⁺¹/n+1)

V = 4t³/3 + 2t²/2 + 2t

V = 4t³/3 + t² + 2t

Now substitute the limits,

v - 2 = [(4 × 5³)/3 + 5² + 2 × 5] - [ (4 × 0)/3 + 0² + 2 ×0]

v - 2 = ( 4 × 125)/3 + 25 + 10

v - 2 = 500/3 + 35

v = 166.67 + 37

v = 203.67 m/s

So,the velocity after t = 5 seconds will be 203.67 m/s.

We basically converted the question into a definite integration question,not so good at LaTeX,so you gotta bear with this messy answer xD

Anyways what else can be done is you can treat the question as an indefinite integration by not putting the values given in the question, instead finding the constant c and then using that value to find the acceleration by substituting it's value in the expression of velocity which you will receive after integration. (re check the solution, late night answering isn't my cup of tea)