Question

The displacement x of a particle verich with the time t as x=14t^-15t+25.Find the position velocity and acceleration of the particle at t=0,when will the velocity of the particle become 0. can we call the motion of the particle as 1 with uniform acceleration.​

Answer 1

Answer:

$x = 25 \\ v = - 15 \\ a = 8 \\ \\ at \: t = \frac{15}{8} \:\\ \: v = 0$

as acceleration is constant

therefore motion is with uniform acceleration.

Answer 2

Answer:

At t = 0,

• position, x = 25 m
• velocity, v = -15 m/s
• acceleration, a = 28 m/s²

Velocity becomes zero at t = 15/28

Step-by-step explanation:

Given :

The displacement x of a particle varies with the time t as x = 14t² - 15t + 25

To find :

• the position velocity and acceleration of the particle at t = 0
• when the velocity of the particle becomes 0
• if the motion of the particle as one with uniform acceleration.​

Solution :

x = 14t² - 15t + 25

At t = 0,

x = 14(0)² - 15(0) + 25

x = 0 + 25

x = 25 m

we get velocity by differentiating displacement.

v = dx/dt

$\sf v=\dfrac{d}{dt}(14t^2-15t+25) \\\\ \sf v=\dfrac{d}{dt}(14t^2)+\dfrac{d}{dt}(-15t)+\dfrac{d}{dt}(25) \\\\ \sf v=28t-15+0 \\\\ \sf v=28t-15$

At t = 0,

velocity, v = 28(0) - 15

v = -15 m/s

And we get acceleration by differentiating velocity,

 a = dv/dt

$\sf a=\dfrac{d}{dt}(28t-15) \\\\ \sf a=\dfrac{d}{dt}(28t)+\dfrac{d}{dt}(-15) \\\\ \sf a=28 \ m/s^2$

At t = 0, acceleration , a = 28 m/s²

Acceleration of the particle is constant as it doesn't vary with time.

Hence, we can call the motion of the particle is with uniform acceleration.

Let's assume at t = t, the velocity is zero.

v = 28t - 15

0 = 28t - 15

28t = 15

t = 15/28 sec