The velocity as a function of time for a car on an amusement park ride is given as v = At2 + Bt with constants A = 2.9m/s3 and B = 1.3m/s2. If the car starts at the origin, what is its position at t = 4.2s?
Well, if the car starts at the origin, we can find its position by integrating the given velocity function. Let's put on our mathematician hats, or should I say, clown wigs!
Taking the integral of the velocity function v = At^2 + Bt, we get the position function:
s = (A/3)t^3 + (B/2)t^2 + C
Now, to find C, the constant of integration, we need to know the initial position of the car. Since it starts at the origin (which we'll define as s = 0 at t = 0), we can set C = 0.
So, the position function becomes:
s = (A/3)t^3 + (B/2)t^2
Plugging in the values A = 2.9 m/s^3, B = 1.3 m/s^2, and t = 4.2 s into the position function, we can calculate the car's position:
s = (2.9/3)(4.2)^3 + (1.3/2)(4.2)^2
Now, to avoid some number soup, let's simplify this expression.
s = 16.2748 + 11.7584
Drum roll, please!
s = 28.0332 meters
So, at t = 4.2 seconds, the car's position is approximately 28.0332 meters from the origin. Hopefully, it didn't get stuck in a clown car traffic jam along the way!
To find the position of the car at t = 4.2s, we need to integrate the velocity function with respect to time.
Given that the velocity function is v = At^2 + Bt, we can find the position function by integrating it.
Integrating v = At^2 + Bt with respect to t, we get:
∫v dt = ∫(At^2 + Bt) dt
Integrating each term separately, we have:
∫v dt = ∫At^2 dt + ∫Bt dt
∫v dt = A∫t^2 dt + B∫t dt
∫v dt = A(t^3/3) + B(t^2/2) + C
Now, let's substitute the values of A, B, and C:
∫v dt = (2.9m/s^3)(t^3/3) + (1.3m/s^2)(t^2/2) + C
To find the constant of integration, we need additional information about the initial conditions. In this case, it is mentioned that the car starts at the origin, which means its initial position (t=0) is 0m.
Using this information, we can set the position function at t = 0 equal to 0:
0 = (2.9m/s^3)(0^3/3) + (1.3m/s^2)(0^2/2) + C
0 = 0 + 0 + C
C = 0
Now, substituting this value of C back into the position function:
∫v dt = (2.9m/s^3)(t^3/3) + (1.3m/s^2)(t^2/2) + 0
∫v dt = (2.9/3)(t^3) + (1.3/2)(t^2)
Finally, to find the position at t = 4.2s, we substitute t = 4.2s into the position function:
position at t = 4.2s = (2.9/3)(4.2^3) + (1.3/2)(4.2^2)
Now, let's calculate the position at t = 4.2s using the given values:
position at t = 4.2s = (2.9/3)(74.088) + (1.3/2)(17.64)
position at t = 4.2s = 24.2432 + 11.442
position at t = 4.2s ≈ 35.6852 meters
Therefore, the position of the car at t = 4.2s is approximately 35.6852 meters.
To find the position of the car at time t = 4.2s, we need to integrate the velocity function with respect to time.
Given: v = At^2 + Bt
To find the position function, integrate the velocity function once:
s = ∫(At^2 + Bt) dt
Integrating At^2 with respect to t gives:
s = (A/3)t^3 + (B/2)t^2 + C
where C is the constant of integration.
We are given that the car starts at the origin, which means its initial position at t = 0s is s = 0. Therefore, we can substitute these values into the position function:
0 = (A/3)(0)^3 + (B/2)(0)^2 + C
0 = C
So, the position function becomes:
s = (A/3)t^3 + (B/2)t^2
Substituting the given values of A = 2.9 m/s^3 and B = 1.3 m/s^2 into the position function:
s = (2.9/3)t^3 + (1.3/2)t^2
Now, plug in t = 4.2s to find the position:
s(4.2) = (2.9/3)(4.2)^3 + (1.3/2)(4.2)^2
Calculating this expression will give you the position of the car at t = 4.2s.
v = A t^2 + B t
v = 2.9 t^2 + 1.3 t
then assuming you know how to integrate
x = (2.9/3)t^3 + (1.3/2) t^2 + c
if x = 0 at t = 0, then c = 0
x = 0.967 t^3 + 0.65 t^2
put in t = 4.2