The velocity of an object is found to be 𝑣(𝑡) = 12√𝑡 + 4 meters per second at time 𝑡
seconds. If at time 𝑡 = 1 seconds, the object is located at position 𝑠(1) = 1 meters, find the
position of the object at time 𝑡 = 4 seconds.
s(t) = 8t^(3/2) + 4t + c
use s(1) = 1 to find c
then find s(4)
yes, C = -11, but you do play fast and loose with those = signs ...
To find C, I can do this right?
y=8t^(3/2) + 4t + C = 1 = 8(1^(3/2)) + 4(1) + C = -11
To find the position of the object at time t = 4 seconds, we need to integrate the velocity function over the time interval [1, 4].
The given velocity function is:
v(t) = 12√t + 4
To integrate the velocity function, we will use the power rule of integration, which states that the integral of x^n with respect to x is (1/(n+1)) * x^(n+1) + C, where C is the constant of integration.
Integrating v(t) with respect to t:
∫v(t) dt = ∫(12√t + 4) dt
Applying the power rule:
= (12/2) * (√t)^2 + 4t + C
= 6t + 4t + C
= 10t + C
Now, we need to evaluate the integral with the given endpoints: t = 1 and t = 4.
Evaluating at t = 4:
s(4) = 10(4) + C
To find the constant of integration (C), we can use the given information that the object is located at position s(1) = 1 meter at time t = 1 second. Plugging in these values into the position function s(t), we get:
s(1) = 10(1) + C
1 = 10 + C
C = -9
Now, plugging in the value of C into the position function:
s(4) = 10(4) + (-9)
Calculating:
s(4) = 40 - 9
s(4) = 31 meters
Therefore, the position of the object at time t = 4 seconds is 31 meters.