Question

20. A particle moves along X direction with
constant acceleration. Its velocity after 4s is 18
mis and displacement is 56 m. Find out its initial
velocity and acceleration.
(1) 10 m/s, 2 m/s2
(2) 8 m/s, 3 m/s2
(3) 8 m/s, 2 m/s
(4) 10 m/s, 3 m/s2
form
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monlar​

Answer 1

Answer:

pls mark it as brainliest

Explanation:

ans is attached in the photo

Answer 2

Answer:

$\boxed{\mathfrak{(1) \ 10 \ m/s, \ 2 \ m/s^2}}$

Given:

Final velocity (v) = 18 m/s

Time taken (t) = 4s

Displacement (s) = 56 m

To Find:

$\sf \star$ Initial velocity (u)

$\sf \star$ Acceleration (a)

Explanation:

Acceleration is defined as the rate of change of velocity with time.

$\boxed{ \bold{ a = \frac{\Delta v}{\Delta t} }}$

$\sf \implies a = \frac{v - u}{ t} \\ \\ \sf \implies a = \frac{18 - u}{4} \: \: \: \: \: \: \: \: ...eq_1$

$\sf From \ 3^{rd} \ equation \ of \ motion: \\ \boxed{ \bold{ {v}^{2} = {u}^{2} + 2as}}$

Substituting values of v, s & a from $\sf eq_1$ we get:

$\sf \implies {(18)}^{2} = {u}^{2} + 2( \frac{18 - u}{4} )(56) \\ \\ \sf \implies 324 = {u}^{2} + 2( \frac{18 - u}{4} ) \times 56 \\ \\ \sf \implies 324 = {u}^{2} + 2(18 - u) \times 14 \\ \\ \sf \implies 324 = {u}^{2} + 28(18 - u) \\ \\ \sf \implies 324 = {u}^{2} + 504 - 28u \\ \\ \sf \implies 324 - 504 = {u}^{2} - 28u \\ \\ \sf \implies 180 = {u}^{2} - 28u \\ \\ \sf \implies {u}^{2} - 28u - 180 = 0 \\ \\ \sf \implies {u}^{2} - 18u - 10u - 180 = 0 \\ \\ \sf \implies u(u - 18) - 10(u - 18) = 0 \\ \\ \sf \implies (u - 18)(u - 10) = 0 \\ \\ \sf \implies u = 18 \ m/s \: \: \: or \: \: \: u = 10 \ m/s$

Final and initial velocity of a particle cannot be equal when the particle is under acceleration.

$\therefore$

$\sf \star$ Initial velocity (u) = 10 m/s

Substituting value of u in $\sf eq_1$ we get:

$\sf \implies a = \frac{18 - 10}{4} \\ \\ \sf \implies a = \frac{8}{4} \\ \\ \sf \implies a = 2 \ m/s^2$

$\therefore$

$\sf \star$ Acceleration (a) = 2 m/s²