Question

The acceleration of a particle traveling along the x axis is given by a(t) = −bt, where b = 7.39 m/s3. In addition, we know that at t = 0, the particle is at the position x0 = 5.00 m and has a velocity of v0 = 10.0 m/s. At the time t = 5.00 s, determine the following for this particle. (Indicate the direction with the sign of your answer.)

Answer 1

Answer:

Acceleration of the particle=$-36.95 \mathrm{m} / \mathrm{sec}^{2}$

Velocity of the particle=$-82.375 \mathrm{m} / \mathrm{sec}$

position of the particle=$98.985 \mathrm{m} / \mathrm{sec}$

Explanation:

Given:

$b=7.39 \mathrm{m} / \mathrm{s}^{3}$

t = 0

$X_{0}=5.00 \mathrm{m}$

$V_{0}=10.0 \mathrm{m} / \mathrm{s}$

$X_{0}=5.00 \mathrm{m}$

$V_{0}=10.0 \mathrm{m} / \mathrm{s}$

To find:

The acceleration of particle at t= 5,00 s

Velocity of the particle at t= 5,00 s

Postion of the particle at t= 5,00 s

Solution:

Finding the value of a:

$v(t)=\int a(t) . b(t)=\left(\frac{-b t^{2}}{2}+a\right) m / s e c$

$\left(\frac{-7.39 t^{2}}{2}+a\right)_{0}=10$

$\left(\frac{-b t^{2}+2 a}{2}\right)_{0}=10$

$\left(-b t^{2}+2 a\right)_{0}=2(10)$

$\left(-b t^{2}+2 a\right)_{0}=20$

$\left(-b(0)^{2}+2 a\right)=2(10)$

$2 a=20$

$a=10$

Finding the value of c

$x(t)=\int v(t) \cdot d(t)=\left(\frac{-b t^{3}}{6}+a t+c\right) m / s e c$

$\left(\frac{-b t^{3}}{6}+a t+c\right)_{0}=5$

Substituting t=0

$\left(\frac{-b(0)^{3}}{6}+a(0)+c\right)=5$

c=5

Finding the velocity of  of particle at t=5

$v(t)=\int a(t) . b(t)=\left(\frac{-b t^{2}}{2}+a\right) m / s e c$

$v(5)=\left(\frac{-(7.39)(5)^{2}}{2}+10\right)$

$v(5)=\left(\frac{-(7.39)(25)}{2}+10\right)$

$v(5)=\left(\frac{-(184.75)}{2}+10\right)$

$v(5)=\left(\frac{-(184.75)}{2}+10\right)$

$v(5)=(-92.375+10)$

$v(5)=(-82.375)$

Position of the particle at x=5

$x(t)=\int v(t) \cdot d(t)=\left(\frac{-b t^{3}}{6}+a t+c\right) m$

$x(5)=\left(\frac{-(7.39)(5)^{3}}{6}+(10)(5)+5\right)$

$x(5)=\left(\frac{-(7.39)(25)}{6}+(50)+5\right)$

$x(5)=\left(\frac{-(923.75)}{6}+(50)+5\right)$

$x(5)=(-153.958+(50)+5)$

$x(5)=(-153.958+(55)$

$x(5)=98.985 \mathrm{m}$

The acceleration of the particle t=5

$a(t)=-b t$

$a(5)=-(7.39)(5)$

$a(5)=-36.95 \mathrm{m} / \mathrm{sec}^{2}$