Using the distance function s(t) = t2 + 4t + 30, "s" is in feet and "t" is in seconds, find the velocity function, v(t), and acceleration function, a(t).
II think you mean
t^2 + 4 t + 30
v = 2 t + 4
a = 2
I do not know why you are supposed to know that if you are not taking calculus.
To find the velocity function, v(t), we need to find the derivative of the distance function, s(t), with respect to time, t.
Given s(t) = t^2 + 4t + 30, let's find v(t) by differentiating s(t) with respect to t:
v(t) = d/dt (t^2 + 4t + 30)
To do this, we can differentiate each term separately:
v(t) = d/dt (t^2) + d/dt (4t) + d/dt (30)
The derivative of t^2 is 2t:
v(t) = 2t + d/dt (4t) + d/dt (30)
The derivative of 4t is 4:
v(t) = 2t + 4 + d/dt (30)
The derivative of a constant term (30) is zero:
v(t) = 2t + 4 + 0
Simplifying, we have:
v(t) = 2t + 4
Therefore, the velocity function, v(t), is 2t + 4.
Next, let's find the acceleration function, a(t), which is the derivative of the velocity function, v(t):
a(t) = d/dt (2t + 4)
To find the derivative, we differentiate each term separately:
a(t) = d/dt (2t) + d/dt (4)
The derivative of 2t is 2:
a(t) = 2 + d/dt (4)
The derivative of a constant term (4) is zero:
a(t) = 2 + 0
Simplifying, we have:
a(t) = 2
Therefore, the acceleration function, a(t), is simply 2.
To find the velocity function, we need to differentiate the distance function, s(t), with respect to time, t.
Given: s(t) = t^2 + 4t + 30
Now, differentiate s(t) with respect to t to find v(t):
To differentiate t^2, we use the power rule: d/dt (t^n) = n*t^(n-1).
So, d/dt (t^2) = 2t.
To differentiate 4t, we use the constant multiple rule: d/dt (c*t) = c*(d/dt)(t) = c*1 = c.
So, d/dt (4t) = 4.
To differentiate constant 30, we use the constant rule: d/dt (c) = 0, where c is a constant.
So, d/dt (30) = 0.
Adding these differentiations together, we get: v(t) = 2t + 4.
Thus, the velocity function is v(t) = 2t + 4.
To find the acceleration function, we need to differentiate the velocity function, v(t), with respect to time, t.
Using the velocity function: v(t) = 2t + 4
The derivative of 2t with respect to t is 2, as t does not have a power.
The derivative of the constant 4 with respect to t is 0 since it is a constant.
So, a(t) = 2.
Hence, the acceleration function is a(t) = 2.