A particle's acceleration is described by the function a(t) = (12 t -30) m/s?, where tis in s. Its initial conditions are xo = 0 m and vo = 0 m/s at t = 0 s. a. At what time isthe velocity is 3m/s? b. What is the particle's position at that time?%3D

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Description: A particle's acceleration is described by the function a(t) = (12 t -30) m/s?, where tis in s. Its initial conditions are xo = 0 m and vo = 0 m/s at t = 0 s. a. At what time isthe velocity is 3m/s? b. What is the particle's position at that time?%3D

Answered: A particle's acceleration is described…

University Physics Volume 1

18th Edition

ISBN:9781938168277

Author:William Moebs, Samuel J. Ling, Jeff Sanny

Publisher:William Moebs, Samuel J. Ling, Jeff Sanny

Chapter3: Motion Along A Straight Line

Section: Chapter Questions

Problem 48P: A particle has a contant acceleration of 6.0m/s2 . (a) If its initial velocity is 2.0 m/s, at what...

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A particle's acceleration is described by the function a(t) = (12 t -30) m/s?, where t
is in s. Its initial conditions are xo = 0 m and vo = 0 m/s at t = 0 s. a. At what time is
the velocity is 3m/s? b. What is the particle's position at that time?
%3D

Expert Solution

Step 1 finding the time at which velocity is 3m/s

The particle's acceleration is described by

$a (t) = (12 t - 30) m / s^{2}$ --------(1)

Given initial conditions are

$x_{0} = 0 m a n d v_{0} = 0 m / s a t t = 0 s$

Since we know that the acceleration is given by

$a = \frac{d v}{d t}$

$d v = a d t$

Integrating both sides

$\int d v = \int a d t + C$ where C is the constant of integration

$v (t) = \int (12 t - 30) d t + C$

$v (t) = 12 \frac{t^{2}}{2} - 30 t + C$ Here we use $\int a t^{n} d t = a \frac{t^{n + 1}}{n + 1} a n d \int a d t = a t$

$v (t) = 6 t^{2} - 30 t + C$ --------(2)

From initial condition, we know at $t = 0, v_{0} = 0$

Using initial condition in (2), we get

$0 = 6 \times 0^{2} - 30 \times 0 + C \Rightarrow C = 0$

Therefore, equation 2 becomes

$v (t) = 6 t^{2} - 30 t$ --------(3)

Let us find the find the time at which v=3m/s. Substitute v=3m/s in (3) and solve t

$3 = 6 t^{2} - 30 t \Rightarrow 6 t^{2} - 30 t - 3 = 0$

This is a quadratic equation in t

$t = \frac{- (- 30) \pm \sqrt{{(- 30)}^{2} - 4 \times 6 \times (- 3)}}{2 \times 6} = \frac{30 \pm \sqrt{900 + 72}}{12} = \frac{30 \pm \sqrt{972}}{12} = \frac{30 \pm 31.18}{12} t = \frac{30 + 31.18}{12} o r t = \frac{30 - 31.18}{12} t = \frac{61.18}{12} = 5.1 s o r t = \frac{- 1.18}{12} = - 0.1 s$ for solving quadratic equation we use $a x^{2} + b x + c = 0, t h e n x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$

Since the time cannot be negative, we neglect the negative value of t

$t = 5.1 s$

Therefor, the particle's velocity is 3m/s at 5.1s.

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