Question

starting from rest a particle moves in a straight line with acceleration a= { 2 + t-2 } m/s^2 then velocity of particle at the end of 4s will be

Answer 1

Answer:answer is 12m/s as shown in picture thish might help u

Explanation:

Answer 2

Answer:

Velocity of particle at the end of 4s will be 12ms⁻¹.

Explanation:

Given:

Acceleration, a=2+|t-2| m/s²

• Acceleration =rate of change of velocity=dv/dt

From given data we understand that,

$|t-2|=+(t-2) for t > 2\\|t-2|=-(t-2) for t \leq 2$

$For t \leq 2$

$a=2-(t-2)\\a=4-t$

$\frac{d v}{d t}=4-t \\\\&\int_{0}^{v_{2}} d v=\int_{0}^{2}(4-t) d t \\\\&v_{2}=\left[4 t-\frac{t^{2}}{2}\right]_{0}^{2} \\\\&=4 \times 2-2 \\&=6 \\\\&\text { For } t > 2 \\\\&a=2+t-2=t \\\\&\frac{d v}{d t}=t\end{aligned}$

$&\int_{\mathrm{v}_{2}}^{\mathrm{v}_{4}} \mathrm{dv}=\int_{2}^{4} \mathrm{tdt}\\ =\left[\frac{\mathrm{t}^{2}}{2}\right]_{2}^{4}\\\frac{4^{2}- 2^{2}}{2} = 6 \\\\ That means \\&\mathrm{v}_{4}-\mathrm{v}_{2}=6 (V_2=6m/s)\\&\mathrm{v}_{4}=\mathrm{v}_{2}+6 \\&\mathrm{v}_{4}=\mathrm{v}_{2}+6 \\&=6+6=12 \mathrm{~m} / \mathrm{s}\end{aligned}$

So velocity at 4s is 12ms⁻¹