Question

2]
8.
The position of a particle moving along the y-axis
is given as y = 3t – t^2, where y is in metre and
t is in second. The time when the particle attains
maximum position in positive y direction will be
[NCERT Pg. 49]
(1) 1.5s
(2) 4.5
(3) 2s
(4) 3 s
f
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Answer 1

Given:

Displacement vs time function has been provided as follows :

y = 3t - t²

To find:

Time at which max displacement in the positive y axis is seen

Concept:

For max displacement , we can differentiate the equation and get its 1st order derivative be equal to zero. That would give us the time at which maxima is reached. And then 2nd order derivative should be less than 0

Calculation:

$y = 3t - {t}^{2}$

$= > \dfrac{dy}{dt} = 3 - 2t$

For maxima , we can say :

$= > 3 - 2t = 0 \\ = > t = \frac{3}{2} = 1.5 \: sec$

Now, 2nd order derivative must be lesser than zero (negative) for maxima .

$= > \dfrac{ {d}^{2}y }{d {t}^{2} } = - 2 \: < 0$

Hence maxima is reached at t = 1.5 secs.

So final answer :

Time for max position = 1.5 secs

Answer 2

Answer:

hope you understand

Explanation:

ans 1 was correct✔