Practice Problems - In Physics Abhay Kumar Pdf

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$= 6t - 2$

$0 = (20)^2 - 2(9.8)h$

A particle moves along a straight line with a velocity given by $v = 3t^2 - 2t + 1$ m/s, where $t$ is in seconds. Find the acceleration of the particle at $t = 2$ s. practice problems in physics abhay kumar pdf

A body is projected upwards from the surface of the earth with a velocity of $20$ m/s. If the acceleration due to gravity is $9.8$ m/s$^2$, find the maximum height attained by the body.

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Given $u = 20$ m/s, $g = 9.8$ m/s$^2$

At $t = 2$ s, $a = 6(2) - 2 = 12 - 2 = 10$ m/s$^2$

Using $v^2 = u^2 - 2gh$, we get

Given $v = 3t^2 - 2t + 1$

$\Rightarrow h = \frac{400}{2 \times 9.8} = 20.41$ m

You can find more problems and solutions like these in the book "Practice Problems in Physics" by Abhay Kumar.

At maximum height, $v = 0$

Acceleration, $a = \frac{dv}{dt} = \frac{d}{dt}(3t^2 - 2t + 1)$