Question

object move with velocity 2 m/s & it's acceleration is given by 6t^2_2t+3 find velocity when time is 3s

Answer 1

Answer:

209 m/s

Explanation:

u=2m/s

t=3s

a=6t²-2t+3

At t=3s, a= 6×3²-2×3+3 = 6×9-6+3 =69 m/s²

v=u+at = 2+69×3 = 2+207=209 m/s

Answer 2

Answer:

$\boxed{\mathfrak{Velocity \ at \ time \ 3s = 56 \ m/s}}$

Explanation:

Function of acceleration (a) w.r.t time (t) is given as:

a = 6t² - 2t + 3

Accelration is rate of change of velocity i.e.

$\rm a = \dfrac{dv}{dt}$

So,

$\rm \implies \frac{dv}{dt} = 6 {t}^{2} - 2t + 3 \\ \\ \rm \implies \int\limits^{v_{f}}_{v_i}dv =\int\limits^{t_f}_{t_i} (6 {t}^{2} - 2t + 3)dt$

Initial velocity ($\rm v_i$) = 2 m/s

Initial time ($\rm t_i$ )= 0 s

Final time ($\rm t_f$ )= 3 s

$\rm \implies \int\limits^{v_{f}}_{2}dv =\int\limits^{3}_{0} (6 {t}^{2} - 2t + 3)dt \\ \\ \rm \implies v\Big|_2^{v_{f}} = 2 {t}^{3} - {t}^{2} + 3t\Big|_0^{3} \\ \\ \rm \implies v_{f} - 2 = 2 {(3)}^{3} - {3}^{2} + 3(3)\\ \\ \rm \implies v_{f} - 2 = 54 - 9 + 9\\ \\ \rm \implies v_{f} = 54 + 2\\ \\ \rm \implies v_{f} = 56 \: m {s}^{ - 1}$

Final velocity ($\rm v_{f}$) = 56 m/s