A particle is moving along the x-axis so that its position at t ≥ 0 is given by s(t)=(t)In(2t). Find the acceleration of the particle when the velocity is first zero.
a)2e^2
b)2e
c)e
d)None of these
Any help is greatly appreciated
To find the acceleration of the particle when the velocity is first zero, we need to follow these steps:
Step 1: Find the velocity function by taking the derivative of the position function with respect to time.
s(t) = t * ln(2t)
To take the derivative, we can use the product rule.
s'(t) = (t * d(ln(2t))/dt) + (ln(2t) * dt/dt)
The derivative of ln(2t) can be calculated using the chain rule as:
d(ln(2t))/dt = 1/(2t) * 2
d(ln(2t))/dt = 1/t
Now, simplifying s'(t):
s'(t) = t/t + ln(2t)
s'(t) = 1 + ln(2t)
Step 2: Set the velocity function equal to zero and solve for t.
1 + ln(2t) = 0
ln(2t) = -1
Using the logarithmic property, we can rewrite this as:
2t = e^(-1)
t = (1/2)e^(-1)
Step 3: Find the acceleration function by taking the derivative of the velocity function.
a(t) = d(s'(t))/dt
a(t) = d(1 + ln(2t))/dt
The derivative of 1 with respect to t is zero since it is a constant.
a(t) = d(ln(2t))/dt
Using the chain rule, the derivative of ln(2t) is:
d(ln(2t))/dt = 1/t
So, the acceleration function is:
a(t) = 1/t
Step 4: Find the acceleration when t = (1/2)e^(-1).
Substituting t = (1/2)e^(-1) into the acceleration function:
a(t) = 1/[(1/2)e^(-1)]
a(t) = 2e
Therefore, the acceleration of the particle when the velocity is first zero is 2e.
Answer: b) 2e
To find the acceleration of the particle when the velocity is first zero, we need to follow several steps:
Step 1: Determine the velocity function from the given position function.
The velocity function is the derivative of the position function with respect to time (t). Thus, we have:
v(t) = d/dt[s(t)]
Step 2: Find the time (t) when the velocity is first zero.
This occurs when v(t) = 0.
Step 3: Calculate the acceleration of the particle at the time when the velocity is first zero.
The acceleration function can be obtained by taking the derivative of the velocity function, defined as:
a(t) = d/dt[v(t)]
Step 4: Solve the acceleration function for the time (t) when the velocity is zero to find the value of acceleration (a).
Let's proceed with these steps:
Step 1: Determine the velocity function.
Taking the derivative of s(t) with respect to t, we have:
v(t) = d/dt[(t)ln(2t)]
To simplify this, we can use the product rule of differentiation:
v(t) = t * d/dt[ln(2t)] + ln(2t) * d/dt[t]
The derivative of ln(2t) with respect to t can be found using the chain rule as 1/t.
Therefore, we have:
v(t) = t * (1/t) + ln(2t) = 1 + ln(2t)
Step 2: Find the time (t) when the velocity is first zero.
To find when v(t) = 0, we set 1 + ln(2t) = 0 and solve for t:
1 + ln(2t) = 0
ln(2t) = -1
2t = e^(-1)
2t = 1/e
t = 1/(2e)
Step 3: Calculate the acceleration of the particle at the time when the velocity is first zero.
Now, we need to find the acceleration function a(t) by taking the derivative of v(t):
a(t) = d/dt[1 + ln(2t)]
a(t) = (d/dt[1]) + (d/dt[ln(2t)])
Since the derivative of a constant (1) with respect to t is zero and the derivative of ln(2t) with respect to t is 1/t, we have:
a(t) = 0 + 1/t
a(t) = 1/t
Step 4: Solve the acceleration function for the time (t) when the velocity is zero.
Substitute t = 1/(2e) into a(t):
a(t) = 1/(1/(2e))
a(t) = 2e
Therefore, the acceleration of the particle when the velocity is first zero is 2e.
Answer: a) 2e^2
v =ds/dt =d/dt [t ln 2t ]
= t(2/2t) + ln 2t = ln 2t
This is undefined at t = 0
a = dv/dt = 1/t
I would pick none of above