Asked by Indira
                PLEASE HELP!!
The x and y cooridinates of an object are given by
x(t) = (10 m/s)t + (5 m/s2)t2 + (3 m/s3)t3
y(t) = (20 m/s)t - (10 m/s2)t2 + (2 m/s3)t3
where x and y are in meters and t is in seconds. What is the object's acceleration at t = 1 s?
Please explain in detail how to do this problem.
Thankss
            
        The x and y cooridinates of an object are given by
x(t) = (10 m/s)t + (5 m/s2)t2 + (3 m/s3)t3
y(t) = (20 m/s)t - (10 m/s2)t2 + (2 m/s3)t3
where x and y are in meters and t is in seconds. What is the object's acceleration at t = 1 s?
Please explain in detail how to do this problem.
Thankss
Answers
                    Answered by
            drwls
            
    You do the first part by plugging in 1 for t and computing x and y. 
You should use ^ marks to denote exponents in your equations. You can leave out the dimensions with the undertanding that t is in seconds and distance s in meters.
x(t) = 10 t + 5 t^2 + 3 t^3
y(t) = 20 t -10 t^2 + 2 t^3
For the acceleration, take the second derivative of x(t) and y(t)
x"(t) = 10 + 18t
y"(t) = -20 + 12t
Once again, plug in t = 1 for the two components of acceleration at that time.
    
You should use ^ marks to denote exponents in your equations. You can leave out the dimensions with the undertanding that t is in seconds and distance s in meters.
x(t) = 10 t + 5 t^2 + 3 t^3
y(t) = 20 t -10 t^2 + 2 t^3
For the acceleration, take the second derivative of x(t) and y(t)
x"(t) = 10 + 18t
y"(t) = -20 + 12t
Once again, plug in t = 1 for the two components of acceleration at that time.
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