a car is at a distance S, in miles, from its starting point in T hours, given by s(t)=10t^2
a).find s(2) and s(5).
b).find s(5)-s(2). what does this represent
c).find the average rate of change from t=2 to t=5. this id the average velocity
A: s(2) = 10(2^2) = 40 s(5) = 10(5^2) = 250 B: 250-40 = 210miles, this represents the distance traveled from the 2nd to 5th hour C: Use the average rate of change formula: (250-40)/(5-2) = 210/3 = 70mph
