![]() ![]() And so all this tells you is what the height is at different times. The pumpkin is shot straight up into the air. It’s important to remember that a graph like this does not necessarily indicate the path of the pumpkin. Actually I should change this, it should be time t. I’ve drawn a graph of the position the height if the pumpkin over time. Now let’s take a look at how we interpret this average velocities on a graph. So you can see that the pumpkin is slowing down a little bit. So 68 divided by 2 is 34 feet per second. This is going to give me 68 feet over 2 seconds. F(4) is 268 and f(2) is 200 still, then we have 2 again in the denominator. Remember it's final position minus initial over the change in time. Now what about between 2 and 4? The change in position here is f of 4 minus f(2). So this is going to give me 98 feet per second. So 200 minus 4 is 196 and that’s in feet. ![]() And f(0) is 4 and the change in time is 2. So I need to calculate the change in position which is f(2) minus f(0), over the change in time which is 2 minus 0. Let’s start with the interval between 0 and 2. And part says compute the average of velocities on the interval t between 0 and 2 and t between 2 and 4. Here’s my table of values for f(t) and for time. A pumpkin is launched into the air time t is in seconds, and height of the pumpkin, f(t) is in feet. ![]() If the man walks around in a circle and comes back to the same point where he started in a circle, then the change in his position is zero and the displacement is also zero.Let’s compute some average velocities. He has completed two rounds around the rectangle and now he is at the starting point.Ī man starts walking from a point on a circular field of radius 0.5 km and 1 hour later he finds himself at the same point where he initially started. ![]() Runner runs around the rectangle twice and the distance covered : The perimeter of the rectangle is the distance travelled in one round. If the total time he takes to run around the track is 100 seconds, determine average speed and average velocity. He travels around rectangle track twice, finally running back to starting point. What is his average velocity?Ī runner is running around rectangle track with length = 50 meters and width = 20 meters. He then reverses and drives 12 km back down the road in 3 minutes. Determine average velocity.Ī truck driver drives 20 km down the road in 5 minutes. The average velocity between t = 1 and t = 2 is given byĪ car travels along a straight road to the east for 120 meters in 4 seconds, then go to the west for 40 meters in 1 second. What is the average velocity between t = 1 and t = 2 seconds? Where t is measured in seconds (we are neglecting air resistance). The average velocity between t = 0 and t = 10 is given byĪn object is dropped from the observation deck of the CN tower so that its height in meters is given by The position of a car is given by s = 10 + 5t + 20t 2 meters at t seconds. What is the average velocity between t = 0 and t = 10 seconds? The average velocity between t = 2 and t = 5 is given by Let s 1 be the position of an object at time t 1 and s 2 be the position of the same object time t 2, then the average velocity over the time interval (t 2 - t 1 ) is defined byĭisplacement : Distance between the starting and ending point.Īn object moves along a straight line so that its position in meters is given by s(t) = t 3 - 6t 2 + 9t for all time in t seconds. Find the average velocity of the object between t = 2 and t = 5 seconds. ![]()
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