# A particle is moving along a line according to the equation of motion s = 0.5 t^2 + 4t / (t + 1), where s is in meters and t is in seconds. What is the instantaneous velocity when the acceleration is zero?

## s = 0.5 t^2 + 4t / (t + 1)

v = ds/dt = t + ( (t+1)(4) - 4t(1) )/(t+1)^2
= t + 4/(t+1)^2

a = dv/dt = 1 -8(t+1)^-3 = 1 - 8/(t+1)^3
= 0
1 = 8/(t+1)^3
(t+1)^3 = 8
t + 1 = 2
t = 1 s
then when t = 1,
v = 1 + 4/2^2 = 2 m/s

## v = ds/dt = t + 4/(t+1)^2

a = 1 - 8/(t+1)^3
a=0 when t=1
v(1) = 2

## To find the instantaneous velocity when the acceleration is zero, we need to first find the acceleration function by taking the derivative of the equation for the velocity, and then finding the value of time (t) when the acceleration is zero.

Let's start by finding the velocity function by taking the derivative of the equation for the position (s) with respect to time (t):

s = (0.5t^2 + 4t) / (t + 1)

To find the derivative, we can use the quotient rule:

ds/dt = [(0.5t^2 + 4t)(1) - (t + 1)(t^2 + 4t)] / (t + 1)^2

Simplifying the equation:

ds/dt = [0.5t^2 + 4t - (t^3 + 5t^2 + 4t)] / (t + 1)^2

ds/dt = (-t^3 - 4t^2 + 4t) / (t + 1)^2

Now, let's find the acceleration function by taking the derivative of the velocity function:

d^2s/dt^2 = (d^2s/dt^2) = (d/dt)(-t^3 - 4t^2 + 4t) / (t + 1)^2

To find the derivative, we can use the quotient rule and simplify the expression:

d^2s/dt^2 = [(-3t^2 - 8t + 4)(t + 1)^2 - (t^3 + 4t^2 - 4t)(2(t + 1))] / (t + 1)^4

d^2s/dt^2 = [(-3t^2 - 8t + 4)(t + 1) - (2t^4 + 10t^3 + 12t^2 - 4t)] / (t + 1)^4

Simplifying the equation further:

d^2s/dt^2 = [-5t^3 - 7t^2 + 2] / (t + 1)^4

To find when the acceleration is zero, we set the acceleration function equal to zero:

0 = -5t^3 - 7t^2 + 2

Now, we need to solve this equation to find the value(s) of time (t) when the acceleration is zero. However, this is a complex mathematical problem, and it might not have an exact solution. If you provide specific values for time (t), I can help you calculate the instantaneous velocity.

## To find the instantaneous velocity when the acceleration is zero, we need to first find the acceleration and then differentiate the equation of motion with respect to time (t).

Given that the equation of motion is:

s = 0.5t^2 + 4t / (t + 1)

To find the acceleration, we need to take the second derivative of the equation of motion with respect to time (t). Let's do that step by step:

1. Differentiate the equation of motion once to find the expression for velocity:
v = ds/dt

To differentiate, we can use the quotient rule to differentiate the numerator and the denominator separately:

For the numerator (0.5t^2 + 4t):
d/dt (0.5t^2 + 4t) = (1 * 0.5t^2) + (t * 1) = 0.5t^2 + t

For the denominator (t + 1):
d/dt (t + 1) = 1

Therefore, the velocity (v) is given by:
v = (0.5t^2 + t) / 1 = 0.5t^2 + t

2. Now, differentiate the velocity equation to find the expression for acceleration:
a = dv/dt

Again, using the quotient rule to differentiate the numerator and the denominator separately:

For the numerator (0.5t^2 + t):
d/dt (0.5t^2 + t) = (1 * 0.5t^2) + (t * 1) = 0.5t^2 + t

For the denominator (1):
d/dt (1) = 0

Therefore, the acceleration (a) is given by:
a = (0.5t^2 + t) / 0 = undefined (since division by zero is undefined)

Since the acceleration is undefined when the denominator is zero, we need to find the value of t when (t + 1) = 0.

t + 1 = 0
t = -1

Now that we have found the value of t when the acceleration is zero, we can substitute this value into the equation for velocity to find the instantaneous velocity:

v = 0.5t^2 + t

Substituting t = -1:

v = 0.5(-1)^2 + (-1) = 0.5 + (-1) = -0.5

Therefore, the instantaneous velocity when the acceleration is zero is -0.5 m/s.