The velocity of a particle at t-0 at origin is u-4-3- m-sec and a constant acceleration is a-6-4- m-sec2- Find the displacement of the particle at t-2 sec

nbsp; nbsp;u=4î+3ĵm/sec a=6î+4ĵm/sec^2= constant Time nbsp;t=2 sec Then displacement is given by nbsp; s=ut +12at^2 =(4î+3ĵ)× 2 + 12(6î+4ĵ)× 4 =(20î+14 ...

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Standard XII

Physics

Acceleration in 1D

The velocity ...

Question

The velocity of a particle at $t = 0$ at origin is $\to u = 4^i + 3^j m / s e c$ and a constant acceleration is $\to a = 6^i + 4^j m / s e c^{2}$ . Find the displacement of the particle at $t = 2 s e c$

(20^i + 14^j)

m

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(20^i - 14^j)

m

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(- 20^i + 14^j)

m

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(- 20^i - 14^j)

m

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Solution

The correct option is A $(20^i + 14^j)$ $m$
$\to u = 4^i + 3^j$ $m / s e c$
$\to a = 6^i + 4^j$ $m / s e c^{2} = constant$
Time $t = 2 s e c$
Then displacement is given by
$\to s = \to u t + \frac{1}{2} \to a t^{2}$
$= (4^i + 3^j) \times 2 + \frac{1}{2} (6^i + 4^j) \times 4$
$= (20^i + 14^j) m$

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The velocity of a particle at t=0 at origin is →u=4^i+3^j m/sec and a constant acceleration is →a=6^i+4^j m/sec2. Find the displacement of the particle at t=2 sec

The correct option is A (20^i+14^j) m →u=4^i+3^j m/sec →a=6^i+4^j m/sec2=constant Time t=2 sec Then displacement is given by →s=→ut+12→at2 =(4^i+3^j)×2+12(6^i+4^j)×4 =(20^i+14^j) m

The velocity of a particle at $t = 0$ at origin is $\to u = 4^i + 3^j m / s e c$ and a constant acceleration is $\to a = 6^i + 4^j m / s e c^{2}$ . Find the displacement of the particle at $t = 2 s e c$

The correct option is A $(20^i + 14^j)$ $m$
$\to u = 4^i + 3^j$ $m / s e c$
$\to a = 6^i + 4^j$ $m / s e c^{2} = constant$
Time $t = 2 s e c$
Then displacement is given by
$\to s = \to u t + \frac{1}{2} \to a t^{2}$
$= (4^i + 3^j) \times 2 + \frac{1}{2} (6^i + 4^j) \times 4$
$= (20^i + 14^j) m$