Question

Pls solve fast .....If a velocity of a particle is given by V= 10+ 2t^2 m/s . The average acceleration between 2 and 5 sec is
Useless answers will be reported

Answer 1

Answer:

Given ----->

velocity of particle ( v ) = 10 + 2t^2 m/s

To find average acceleration,

Average acceleration =

change in velocity / change in time =

V5 - V2 / 5-2

= V5 - V2 / 3 .............(1)

Now, Putting value of t = 5s.

V5 = 10 + 2 (5 ) ^2 m/s

V5 = 10 + 50 m/s = 60 m/s.

Now, Putting value of t = 2s.

V2 = 10 + 2(2)^2 m/s

V2 = 10 + 8 m/s = 18 m/s.

Putting values of V2 and V5 in eq. ( 1)

=

Average acceleration = ( 60 - 18 / 3 ) m/s

= 42 / 3 m/ s

= 14

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Answer 2

$\large{\underline{\underline{\mathfrak{Answer :}}}}$

• Average acceleration between 2 and 5s is 14 m/s²

$\rule{200}{0.5}$

$\underline{\underline{\mathfrak{Step-By-Step-Explanation :}}}$

Given :

• v = 10 + 2t²

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To Find :

• Find acceleration between 2 and 5 s

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Solution :

We have formula :

$\large{\boxed{\sf{a \: = \: \dfrac{dv}{dt}}}} \\ \\ \implies {\sf{a \: = \: \dfrac{d(10 \: + \: 2t^2)}{dt}}} \\ \\ \implies {\sf{a \: = \: 0 \: + \: 2 \: \times \: 2t}} \\ \\ \implies {\sf{a \: = \: 4t}} \\ \\ \LARGE{\underline{\boxed{\sf{a \: = \: 4t \: ms^{-2}}}}}$

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Now, put t = 5

$\implies {\sf{a \: = \: 4(5)}} \\ \\ \implies {\sf{a \: = \: 20}} \\ \\ \large{\boxed{\sf{a_1 \: = \: 20 \: ms^{-2}}}}$

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Now, put t = 2s

$\implies {\sf{a \: = \: 2(4)}} \\ \\ \implies {\sf{a \: = \: 8}} \\ \\ \large{\boxed{\sf{a_2 \: = \: 8 \: ms^{-2}}}}$

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Now, average acceleration be :

$\large{\boxed{\sf{a_{avg} \: = \: \dfrac{a_1 \: + \: a_2}{2}}}} \\ \\ \implies {\sf{a_{avg} \: = \: \dfrac{20 \: + \: 8}{2}}} \\ \\ \implies {\sf{a_{avg} \: = \: \dfrac{28}{2}}} \\ \\ \implies {\sf{a_{avg} \: = \: 14}} \\ \\ \underline{\sf{\therefore \: Average \: acceleration \: between \: 2 \: and \: 5 \: s \: is \: 14 \: ms^{-2}}}$