the position of a particle moving along a coordinate line is s=√(1+4t) , with s in metres and t in seconds. Find the particle's velocity and acceleration at t=6 sec
v = ds/dt = 2/√(1+4t)
a = dv/dt = -4/(1+4t)3/2
now just plug in t=6
Well, isn't that quite a mathematical adventure! Let's get into it, shall we?
To find the particle's velocity, we need to take the derivative of the position function with respect to time, which in this case is t. So, let's differentiate the position function s = √(1+4t):
ds/dt = d/dt (√(1+4t))
Now, the chain rule comes into play here. Applying it, we get:
ds/dt = (1/2)*(1+4t)^(-1/2) * d(1+4t)/dt
The derivative of (1+4t) with respect to t is simply 4, so we can simplify the expression further:
ds/dt = (2/√(1+4t))
Taking the velocity at t = 6 seconds gives us:
v = (2/√(1+4*6))
v = (2/√(1+24))
v = (2/√25)
v = 2/5
So, the particle's velocity at t = 6 seconds is 2/5 m/s.
Now, let's tackle the acceleration. We need to find the derivative of velocity, which is the second derivative of the position function. Let's differentiate the velocity function with respect to time:
dv/dt = d/dt (2/√(1+4t))
Using the chain rule again, we have:
dv/dt = (-2/2)*(1+4t)^(-3/2) * d(1+4t)/dt
Differentiating (1+4t) gives us 4, so we have:
dv/dt = (-1)*(1+4t)^(-3/2) * 4
Simplifying this further, we get:
dv/dt = (-4)/(2√(1+4t))^3
dv/dt = (-2)/(√(1+4t))^3
Taking the acceleration at t = 6 seconds:
a = (-2)/(√(1+4*6))^3
a = (-2)/(√(1+24))^3
a = (-2)/(√25)^3
a = (-2)/(5)^3
a = (-2)/125
a = -0.016
So, the particle's acceleration at t = 6 seconds is approximately -0.016 m/s².
Now, isn't that a mathematical rollercoaster? Hang on tight!
To find the particle's velocity and acceleration, we need to differentiate the position equation with respect to time (t).
Given: s = √(1 + 4t)
Step 1: Differentiate the position equation with respect to time (t) to find the particle's velocity:
v = ds/dt
Step 1.1: Differentiate the square root function using the chain rule:
Let u = 1 + 4t
du/dt = 4
Step 1.2: Apply the chain rule:
v = (1/2)(1 + 4t)^(-1/2) * 4
Step 1.3: Simplify the expression:
v = 2(1 + 4t)^(-1/2)
Step 2: Differentiate the velocity equation with respect to time (t) to find the particle's acceleration:
a = dv/dt
Step 2.1: Differentiate the velocity equation:
a = d/dt [2(1 + 4t)^(-1/2)]
Step 2.2: Apply the chain rule:
a = d/dt [(2)(1 + 4t)^(-1/2)]
= (d/dt)(2)(1 + 4t)^(-1/2)
Step 2.3: Differentiate the function using the chain rule:
Let u = 1 + 4t
du/dt = 4
Step 2.4: Apply the chain rule:
a = (d/dt)(2)(1 + 4t)^(-1/2)
= (2)(-1/2)(d/dt)(1 + 4t)^(-1/2)
= -2(1 + 4t)^(-1/2 - 1) * 4
Step 2.5: Simplify the expression:
a = -8(1 + 4t)^(-3/2)
Therefore, the particle's velocity at t = 6 seconds is v = 2(1 + 4(6))^(-1/2) and its acceleration at t = 6 seconds is a = -8(1 + 4(6))^(-3/2).
To find the velocity and acceleration of the particle, we need to differentiate the position function with respect to time.
Given:
Position function: s = √(1 + 4t)
Step 1: Differentiate the position function to find the velocity function.
To find the velocity, we differentiate the position function s with respect to t.
ds/dt = d(√(1 + 4t))/dt
To differentiate √(1 + 4t), we can apply the chain rule.
Let u = 1 + 4t
Then, √u = √(1 + 4t)
Using the chain rule, d(√u)/dt = (1/2)*(1/√u) * du/dt
du/dt = d(1 + 4t)/dt = 4
Substituting back, we get:
ds/dt = (1/2)*(1/√u) * du/dt = (1/2)*(1/√(1 + 4t)) * 4 = 2/(√(1 + 4t))
So, the velocity function v(t) = ds/dt = 2/(√(1 + 4t))
Step 2: Differentiate the velocity function to find the acceleration function.
To find the acceleration, we differentiate the velocity function v with respect to t.
dv/dt = d(2/(√(1 + 4t)))/dt
To differentiate 2/(√(1 + 4t)), we can use the quotient rule.
Let u = 2 and v = √(1 + 4t)
Then, dv/dt = d(√u)/dt = (1/2)*(1/√u) * du/dt
du/dt = 0 (since u = 2)
Substituting back, we get:
dv/dt = (1/2)*(1/√u) * du/dt = (1/2)*(1/√(1 + 4t)) * 0 = 0
So, the acceleration function a(t) = dv/dt = 0
Now, let's find the velocity and acceleration at t = 6 seconds.
Velocity at t = 6 seconds:
v(6) = 2/(√(1 + 4*6))
= 2/(√(1 + 24))
= 2/(√25)
= 2/5 = 0.4 m/s
Acceleration at t = 6 seconds:
a(6) = 0
Therefore, the particle's velocity at t = 6 seconds is 0.4 m/s and its acceleration is 0.