The position of a particle moving along the x-axis is given by s(t)=2t2+4. Use difference quotients to find the velocity v(t) and acceleration a(t), filling in the following expressions as you do so:
v(t)=limh->0 [_____ / h ] = ____
a(t)=limh->0 [ ____ / h ] = _____
ok my answer for v(t)= h(4t+h)/ h = 4t
and for a(t)= 4h/h = 4
for some reason the first part of v(t) is wrong, could someone please tell me what's wrong with it??
v(t)= h(4t+h)/ h = 4t
This should be,
v(t)= 2h(2t + h)/h = 2(2t + h) = 4t + 2h
v(t) = (2(t + h)^2 + 4 - (2t^2 + 4))/h
v(t)=(2t^2 + 4ht + 2h^2 + 4 - 2t^2 - 4)/h
v(t) = (4ht + 2h^2)/h
v(t) = 2h(2t + h)/h
v(t) = 2(2t + h)
v(t) = 4t + 2h
Then, lim h-->0
v(t) = 4t + 2h
v(t) = 4t + 0 = 4t
Velocity, v is the derivative, f'(t).
The derivative of s(t) = 2t^2 + 4 is,
f'(t) = 4t
So, if your problem answer says v = 4t is wrong, you will have to ask your teacher.
Well, it seems like you made a small mistake in finding the velocity. Let's go through the calculations together:
To find the velocity, we need to take the derivative of the position function with respect to time, which means we differentiate the function s(t) = 2t^2 + 4.
s(t) = 2t^2 + 4
Now, let's differentiate s(t) using the power rule:
v(t) = d/dt (2t^2 + 4)
= 4t
So, the correct expression for velocity is v(t) = 4t.
Regarding acceleration, we need to differentiate the velocity function v(t) = 4t with respect to time:
a(t) = d/dt (4t)
= 4
Therefore, the correct expression for acceleration is a(t) = 4.
I hope that clears things up! Let me know if you have any more questions.
To find the velocity, we need to calculate the difference quotient of the position function with respect to time.
The position function is given by s(t) = 2t^2 + 4.
To calculate the difference quotient, we evaluate the following expression:
v(t) = lim(h->0) [(s(t + h) - s(t)) / h]
Let's substitute the values into the expression:
v(t) = lim(h->0) [(2(t + h)^2 + 4 - (2t^2 + 4)) / h]
Expanding and simplifying:
v(t) = lim(h->0) [(2t^2 + 4t + 2h^2 + 4 - 2t^2 - 4) / h]
v(t) = lim(h->0) [(4t + 2h^2) / h]
Next, we can cancel out one of the t^2 terms as it approaches zero as h approaches zero:
v(t) = lim(h->0) [(4t + 2h^2) / h]
Now we factor out 2 from the numerator:
v(t) = lim(h->0) [2(2t + h^2) / h]
Dividing both terms in the numerator by h:
v(t) = lim(h->0) [2(2t / h + h^2 / h)]
Simplifying:
v(t) = lim(h->0) [4t / h + 2h]
v(t) = 4t + lim(h->0) [2h]
As h approaches zero, the second term, 2h, also approaches zero. So we have:
v(t) = 4t
Therefore, the correct expression for velocity is:
v(t) = 4t
Your answer for v(t) is correct. Now let's calculate the acceleration:
a(t) = lim(h->0) [(v(t + h) - v(t)) / h]
Substituting the velocity function v(t) = 4t:
a(t) = lim(h->0) [((4(t + h)) - (4t)) / h]
a(t) = lim(h->0) [(4t + 4h - 4t) / h]
a(t) = lim(h->0) [4h / h]
As h approaches zero, the term 4h also approaches zero. So we have:
a(t) = 0
Therefore, the correct expression for acceleration is:
a(t) = 0
Your answer for a(t) is correct.
To find the velocity and acceleration of a particle, we need to differentiate the position function with respect to time. Let's start by finding the velocity.
The velocity is defined as the derivative of the position function, which measures how the position changes with respect to time. In this case, the position function is given as s(t) = 2t^2 + 4.
To find the velocity, we differentiate the position function with respect to t:
v(t) = d(s(t))/dt
Now, let's take the derivative. When differentiating a polynomial function like 2t^2, we apply the power rule, which states that for a term of the form ct^n, the derivative is given by nct^(n-1).
For the position function s(t) = 2t^2 + 4, we differentiate each term separately:
d(2t^2)/dt = 2 * (d(t^2)/dt) = 2 * 2t = 4t
Since the derivative of a constant term (4) is zero, it disappears.
Therefore, the velocity function v(t) = 4t.
Now, let's calculate the acceleration.
The acceleration is defined as the derivative of the velocity function, which measures how the velocity changes with respect to time. In this case, the velocity function is v(t) = 4t.
To find the acceleration, we differentiate the velocity function with respect to t:
a(t) = d(v(t))/dt
By differentiating the velocity function, we obtain:
d(4t)/dt = 4 * (d(t)/dt) = 4 * 1 = 4
Therefore, the acceleration function a(t) = 4.
In conclusion, the correct expressions for the velocity and acceleration are:
v(t) = 4t
a(t) = 4
Your calculation for the acceleration is correct, but there seems to be an error in your calculation for the velocity. The correct expression for v(t) is 4t, not h(4t+h)/h.