Question

If the particle is moving along a straight line given by the relation s=2-3t+4t^3 where s is in cm.,and t in sec.,it's average velocity during the third sec is?

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Answer 1

According to the question

A particle is moving along a straight line given by the relation;

s = 2 - 3t + 4t³

Now, we have to find the average velocity, for this we will differentiate s with respect to t;

\begin{lgathered}= > \frac{ds}{dt} = \frac{d(2 - 3t + 4 {t}^{3} )}{dt} \\ \\ = > \frac{d2}{dt} - \frac{d3t}{dt} + \frac{d4 {t}^{3} }{dt} \\ \\ = > 0 - 3 + \frac{12}{2} {t}^{2} \\ \\ = > 6 {t}^{2} - 3\end{lgathered}

=>

dt

ds

=

dt

d(2−3t+4t

3

)

=>

dt

d2

−

dt

d3t

+

dt

d4t

3

=>0−3+

2

12

t

2

=>6t

2

−3

After differentiating ds/dt we get 6t² - 3

Now, we have to find average speed at t = 3 sec

So, we will put the value in ds/dt;

=> 6t² - 3 = 6(3)² - 3

=> 6 × 9 - 3

=> 54 - 3

=> 51

So, average speed at t = 3 is 51 m/s

Answer 2

Explanation:

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