+564 Okay so this is a two part question. I solved the first part but need help solving the second.
The first part goes as:
The height of a certain object is given by h(t)=t2−1. Find the average velocity over the time period [2,5].
Somebody helped me with a similar problem like this before (That awesome person knows who they are ;) ). The answer I got for the one was and average velocity 7
The second part of the question is:
In the previous problem, find the exact value of the instantaneous velocity at t=2. One way is to evaluate the limit with a=2
+118703 Hi chilledz3non,
Well the exact instantaneous velocity is given by h'(t) = 2t so when t=2 velocity=2*2=4 units per time.
Now perhaps you are meant to do it from first principles so lets think about that.
Average velocity between t=some value, I'll call it t1 and t=2 would be f(t1)−f(2)t1−2
The instantaneous velocity would be given by this as t1 tends to 2
That is
limt1→2f(t1)−f(2)t1−2f(t1)=(t1)2−1f(2)=22−1=3$Instantaneousvelocityatt=2is$limt1→2f(t1)−f(2)t1−2=limt1→2(t1)2−1−3t1−2=limt1→2(t1)2−4t1−2=limt1→2((t1)−2)((t1)+2)t1−2=limt1→2(t1−2)(t1+2)t1−2=limt1→2(t1+2)=4
Maybe Alan or Heureka could tell me why my limits are not displaying properly. 
+118703 Hi chilledz3non,
Well the exact instantaneous velocity is given by h'(t) = 2t so when t=2 velocity=2*2=4 units per time.
Now perhaps you are meant to do it from first principles so lets think about that.
Average velocity between t=some value, I'll call it t1 and t=2 would be f(t1)−f(2)t1−2
The instantaneous velocity would be given by this as t1 tends to 2
That is
limt1→2f(t1)−f(2)t1−2f(t1)=(t1)2−1f(2)=22−1=3$Instantaneousvelocityatt=2is$limt1→2f(t1)−f(2)t1−2=limt1→2(t1)2−1−3t1−2=limt1→2(t1)2−4t1−2=limt1→2((t1)−2)((t1)+2)t1−2=limt1→2(t1−2)(t1+2)t1−2=limt1→2(t1+2)=4
Maybe Alan or Heureka could tell me why my limits are not displaying properly.
