Question

38) The acceleration of a moving particle varies directly proportional to time. The particle starts from rest and att = 1s, its position and velocity are 2 m and 2 m/s. Find its equation of motion.​

Answer 1

Answer:

Motion of the particle is given as

$\Delta x = \frac{2t^3}{3} + \frac{4}{3}$

Explanation:

As we know that acceleration varies directly with time

so we have

$a = k t$

so we have

$a = \frac{dv}{dt}$

$\frac{dv}{dt} = kt$

$v = \frac{kt^2}{2} + c$

at t = 0 speed of the particle is zero

so we have

$0 = 0 + c$

so we have

$v = \frac{kt^2}{2}$

also at t = 1 speed v = 2 m/s

$2 = \frac{k}{2}$

k = 4

now we have

$v = \frac{dx}{dt}$

$\frac{dx}{dt} = 2t^2$

so we have

$\Delta x = \frac{2t^3}{3} + c$

now at t = 1 the particle will reach at x = 2

$2 = \frac{2}{3} + c$

$c = \frac{4}{3}$

now motion of the particle is given as

$\Delta x = \frac{2t^3}{3} + \frac{4}{3}$

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Topic : Kinematics

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Answer 2

Motion of the particle is given as

\Delta x = \frac{2t^3}{3} + \frac{4}{3}Δx=32t3+34

Explanation:

As we know that acceleration varies directly with time

so we have

a = k ta=kt

so we have

a = \frac{dv}{dt}a=dtdv

\frac{dv}{dt} = ktdtdv=kt

v = \frac{kt^2}{2} + cv=2kt2+c

at t = 0 speed of the particle is zero

so we have

0 = 0 + c0=0+c

so we have

v = \frac{kt^2}{2}v=2kt2

also at t = 1 speed v = 2 m/s

2 = \frac{k}{2}2=2k

k = 4

now we have

v = \frac{dx}{dt}v=dtdx

\frac{dx}{dt} = 2t^2dtdx=2t2

so we have

\Delta x = \frac{2t^3}{3} + cΔx=32t3+c

now at t = 1 the particle will reach at x = 2

2 = \frac{2}{3} + c2=32+c

c = \frac{4}{3}c=34

now motion of the particle is given as

\Delta x = \frac{2t^3}{3} + \frac{4}{3}Δx=32t3+34

#Learn

Topic : Kinematics

https://brainly.in/question/1376976