As displacement function is the antiderivative of the velocity function, we can find displacement function using the information given in the problem.
i) Antiderivative of 9t is 4.5t^2 + C
ii) Antiderivate of 10sqrt(t) - 1/t is (20/3)t^(3/2) - ln(|t|) + C
We can think about two cases where the time displacement is 10 cm. First case is when time is between 0 to 1 seconds. Second case is when time is greater than 1 second. Let's first consider the first case.
First Case:
s = 4.5t^2 (from 0 to x seconds)
Δs = 4.5t^2 - 0 = 10
t = 1.49 seconds.
As this value of time is outside the domain we setted (0 to 1), we can conclude that there is no solution in first case. So, the answer must be in the second case.
Second Case:
s = 4.5t^2 (from 0 to 1 seconds)
s = (20/3)t^(3/2) - ln(|t|) (from 1 to x seconds).
Δs = 4.5t^2 (from 0 to 1 seconds) + (20/3)t^(3/2) - ln(|t|) (from 1 to x seconds)
Like always, thank you so much!
Hi,
Thank you for your question.
As displacement function is the antiderivative of the velocity function, we can find displacement function using the information given in the problem.
i) Antiderivative of 9t is 4.5t^2 + C
ii) Antiderivate of 10sqrt(t) - 1/t is (20/3)t^(3/2) - ln(|t|) + C
We can think about two cases where the time displacement is 10 cm. First case is when time is between 0 to 1 seconds. Second case is when time is greater than 1 second. Let's first consider the first case.
First Case:
s = 4.5t^2 (from 0 to x seconds)
Δs = 4.5t^2 - 0 = 10
t = 1.49 seconds.
As this value of time is outside the domain we setted (0 to 1), we can conclude that there is no solution in first case. So, the answer must be in the second case.
Second Case:
s = 4.5t^2 (from 0 to 1 seconds)
s = (20/3)t^(3/2) - ln(|t|) (from 1 to x seconds).
Δs = 4.5t^2 (from 0 to 1 seconds) + (20/3)t^(3/2) - ln(|t|) (from 1 to x seconds)
Δs = 4.5(1)^2 - 0 + [ (20/3)(x^(3/2) - ln(|x|) ] - [ (20/3)(1)^(3/2) - ln(|1|) ] = 10
Δs = 4.5 + (20/3)(x)^(3/2) - ln(|x|)- (20/3) = 10
(20/3)(x)^(3/2) - ln(|x|) = 73/6.
Now, you can use a calculator to solve for x.
The calculator tells us that the value for x is 1.528.
Does this answer your question?