Question

Open with A gorilla (wearing a parachute) jumped on or the top of a building. We were able to record the velocity of the gorilla with respect to time twice each second. The data is shown below. Note that he touched the ground just after 5 seconds. Velocity (in feet per second) Time (in seconds) 0 0 0.5 5 1.0 7 1.5 8 2.0 11 2.5 11.5 3.0 12 3.5 نیا 4.0 15.5 4.5 18 5.0 19

Answer 1

1)In the given problem, we have to find the approximate distance the gorilla fell for each half second. Now we know that velocity(v) is nothing but distance(d) per unit time(t) (v=d/t), therefore, distance is velocity into time. That is, d=vt

In the each range of time interval, we can either consider right end point (velocity at right end) and calculate the distance using above formula or we can use left end point(velocity at left end) and calculate the distance using above formula. Note that the time interval is the same for all calculations, which is half second, so t=0.5 s. Let us consider right end point, then the distance gorilla fell for each half second for the first case of 0-0.5s second is

distance (d in feet)= velocity (right end point in feet/second) x time interval (t= 0.5 s) = 5 x 0.5 = 2.5 feet

Similarly we calculate the distance for the entire table

time (s)	v(feet/s)	distance (feet)
0-0.5	5	5 x 0.5 = 2.5
0.5-1.0	7	7 x 0.5 =3.5
1.0-1.5	8	8 x 0.5 = 4
1.5-2.0	11	11 x 0.5 = 5.5
2.0-2.5	11.5	11.5 x 0.5 = 5.75
2.5-3.0	12	12 x 0.5 = 6
3.0-3.5	13	13 x 0.5 =6.5
3.5-4.0	15.5	15.5 x 0.5 = 7.75
4.0-4.5	18	18 x 0.5 = 9
4.5-5.0	19	19 x 0.5 = 9.5

2) As we can see in the last column, the distance traveled by the gorilla goes on increasing as its velocity goes on increasing. And eventually the distance traveled by the gorilla when he fell during the time interval from 4.5-5.0 s is the maximum, 9.5 feet. Thus we can see from completed table 1 that the idea put forth in point 2 makes sense

3) In order to find the total distance that the gorilla fell from the time he jumped of the building until the time he landed on the ground, we add all the distances calculated in the point number 1,

Thus the sum of distance is 2.5+3.5+4+5.5+5.75+6+6.5+7.75+9+9.5 = 60 feet. Therefore, total distance the gorilla fell is 60 feet.

As we have considered the right endpoints while calculating the distance in point #1, the total distance calculated is the right hand sum. Also since the velocity is increasing with time, the distance calculated using the right handed sum is the maximum distance the gorilla can travel while he falls from the building and lands on the ground.

4) The velocity is increasing with time and it is maximum at the end of the interval (as we can see it is 19 feet/s), we have taken right hand sum to solve the problem, due to which the total distance calculated (60 feet) is overestimated. In other words when we take the summation of the distance in table 1, we are basically considering that the velocity is constant throughout the interval and it is equal to the velocity at the end of the interval. But in reality, the velocity is not constant and is increasing with time so the previous velocities are less than the velocity of end interval and thus 60 feet distance is an overestimate.

5) Note that in the given problem, velocity is increasing with time through the interval of 0 to 5 seconds, so

To calculate the overestimated approximation, we take the right hand sum by using the right end point and Δt = 0.5s,

The distance (overestimate) = Δ t x (sum of velocities using right end point) = 0.5 x ( 5+7+8+11+11.5+12+13+15.5+18+19)= 60 feet

Now, to calculate the underestimated approximation, we take the left hand sum by using the left end point and Δ t = 0.5s

The distance (underestimate) =  Δ t x ( sum of velocities using left end point) = 0.5 x ( 0+5+7+8+11+11.5+12+13+15.5+18) = 50.5 feet

Thus OE = 60 feet and UE = 50.f feet

So the gorilla fell at least 50.5 feet

and the gorilla fell at most 60 feet

6) The function of velocity in the form v (t) = v(0) + at, would have helped in providing better approximation. Using the velocity - time curve, it is possible to find the displacement as the area under the curve of the plot of v vs t curve. The time can be divided into small intervals and the corresponding velocity as constant over each interval. In other words we could have taken the time integral of the velocity function for each interval which would have provided the better approximation of the displacement.