Question

For a particle moving along a straight line, velocity, v is given as v = 2t 2 where v is in m/s and time t is in second. The displacement of the particle in the first 2 seconds is

Answer 1

Explanation:

v=ds/dt

(2t^2)dt=ds

after integrating

2t^3/3 =s

put t=2seconds

s=2×2^3/3

=16/3metre

Answer 2

Answer:

$\boxed{\mathfrak{Displacement \ of \ the \ particle \ in \ first \ 2 s = \frac{16}{3} \ m}}$

Given:

Velocity of particle as a function of time:

$\sf v = 2t^2 \ m/s$

To Find:

Displacement of the particle (s) in the first 2 seconds i.e. t = 2 s

Explanation:

As we know:

$\boxed{ \bold{v = \frac{ds}{dt} }}$

$\sf \implies vdt = ds \\ \\ \sf \implies \int\limits^s_0 ds = \int \limits^t_0 vdt \\ \\ \sf v = 2t ^{2} : \\ \sf \implies \int\limits^s_0 ds = \int\limits^{t = 2}_0 2t^{2} \, dt \\ \\ \sf \implies \left[s \right]_0^s = \left[ \frac{ 2 {t}^{3} }{3} \right]_0^2 \\ \\ \sf \implies \left[s - 0 \right] = \left[ \frac{2 {(2)}^{3} }{3} - \frac{2 {(0)}^{3} }{3} \right] \\ \\ \sf \implies s = \left[ \frac{2 \times 8}{3} - 0 \right] \\ \\ \sf \implies s = \frac{16}{3} \: m$

$\therefore$

$\sf Displacement \: of \: the \: particle \: in \: the \: first \: 2 \: seconds = \frac{16}{3} \ m$

Additional information:

$\sf \int\limits^{x_1}_{x_2} x^ndx = \left[ \frac{x^{n + 1}}{n + 1} \right]_{x_2}^{x_1}$