Question

Displacement of a body after time 't' sec is s= 2t² - t + 1 find i) velocity at 2 sec and acceleration to any time 't'​

Answer 1

Given that,

• S = 2t² - t + 1

To Find,

• Velocity and acceleration.

Solution,

$\\\footnotesize\quad\bullet\quad \boxed{\rm v = \dfrac{ds}{dt} }\bigstar$

$\\\footnotesize\quad\bullet\quad \boxed{\rm a = \dfrac{dv}{dt} }\bigstar$

Calculation,

Finding velocity:

$\\\footnotesize\quad\longrightarrow\quad \rm v = \dfrac{d}{dt} \bigg( 2t^{2} - t + 1 \bigg)$

$\\\footnotesize\quad\longrightarrow\quad \rm v = \dfrac{d}{dt} \bigg( 2t^{2}\bigg) - \dfrac{d}{dt}\bigg( t \bigg) + \dfrac{d}{dt} \bigg( 1 \bigg)$

$\\\footnotesize\quad\longrightarrow\quad \rm v = 2 \times 2 (t) - 1 + 0$

$\\\footnotesize\quad\longrightarrow\quad \rm v = 4t - 1 --- (i)$

$\\\footnotesize\quad\longrightarrow\quad \rm v_{(t = 2 s)} = 4(2) - 1$

$\\\footnotesize\quad\therefore\quad \boxed{ \rm v_{(t = 2 s)} = 7 m/s }\bigstar$

Finding Acceleration:

differentiate equation (i) w.r.t time (t)

$\\\footnotesize\quad\longrightarrow\quad \rm a = \dfrac{d}{dt} \bigg(4t - 1 \bigg)$

$\\\footnotesize\quad\longrightarrow\quad \rm a = \dfrac{d}{dt} \bigg(4t \bigg) - \dfrac{d}{dt} \bigg(1 \bigg)$

$\\\footnotesize\quad\longrightarrow\quad \rm a = 4 - 0$

$\\\footnotesize\quad\therefore\quad \boxed{\rm a = 4 m/s^{2}}\bigstar$

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