A car passes by a certain point at time t = 0 hours heading east at 40 miles per hour. At t = 1 hour another car passes by the same point heading north at 50 miles per hour. What is the functiuonthat equals the distance between the two cars?
distance= sqrt[(40t)^2+(50(t-1))^2 ]
check my thinking.
The answer has to be one of the following:
a. L=90t-50
b. L= sqrt(4100t)
c. L= sqrt(4100t^2 +5000t +2500)
d. L= sqrt(4100t^2-5000t + 2500)
e. L= sqrt(90t-50)
And, it is one of those. Can you not do the algebra?
just simplify what bob put you'll get D!
To find the function that equals the distance between the two cars, we can use the concept of relative motion. Let's set up a coordinate system with the initial point where the first car passes by as the origin (0,0).
The position of the first car at any given time can be represented by the equation x = 40t, where x is the eastward distance traveled by the first car and t is the time elapsed in hours.
Similarly, the position of the second car can be represented by the equation y = 50(t - 1), where y is the northward distance traveled by the second car. Note that we subtract 1 from t because the second car starts moving 1 hour after the first car.
To find the distance between the two cars, we can use the distance formula between two points in a coordinate system:
d = sqrt((x - 0)^2 + (y - 0)^2)
Substituting the given equations for x and y, we have:
d = sqrt((40t)^2 + (50(t - 1))^2)
Simplifying the equation further:
d = sqrt(1600t^2 + 2500(t^2 - 2t + 1))
d = sqrt(1600t^2 + 2500t^2 - 5000t + 2500)
d = sqrt(4100t^2 - 5000t + 2500)
Thus, the function that equals the distance between the two cars is d = sqrt(4100t^2 - 5000t + 2500).