Question

Jeremy analyses one of his parachute jumps. He draws a graph showing his velocity up to the opening of his parachute. a) Estimate Jeremy's acceleration at t = 10s to 1 decimal place. b) Estimate his average speed for this part of the jump. Give your answer to two significant figures.

Answer 1

Answer:

Jeremy's acceleration is about $1\,\frac{m}{s^2}$  at t =10 s

His average speed is about 44.5 m/s in this section of his jump approximating with points on the curve (under-estimate)

His average speed is about 46 m/s if using the tangent line (over estimate)

Step-by-step explanation:

Jeremy's acceleration can be estimated by the curve's derivative at that point. That is the slope of the tangent line to the velocity curve at x = 10 sec. Please see attached image where the tangent line is drawn in orange, and the points to use to calculate its slope are drawn in green.

These points are : (6, 42) and (14,50) which using the slope formula give:

$slope=\frac{y_2-y_1}{x_2-x_1}= \frac{50-42}{14-6}=\frac{8}{8} \frac{m}{s^2} = 1\,\frac{m}{s^2}$

So his acceleration at that point is about $1\,\frac{m}{s^2}$

Now, using about the same interval of x-values (from 6 to 14), the corresponding speeds are approximately: 40 (for time 6 seconds) and 49 (for time 14 seconds (look for the red dots on the attached image). Since  the average velocity is given by the integral of the function between those points divided by the length of the interval where it is calculated:

$v_{average}=\frac{area}{interval\,\,length}$

and we don't have the actual velocity function to estimate the integral, we can approximate this area by that of a trapezoid that connects the red dots with the bottom of the horizontal axis (see red trapezoid in the image). Clearly from the image, this approximation would give us an under-estimate of the actual average speed.

The area of this trapezoid is: approximately:

$Trapezoid\,\, area=(49+40)\,8/2=356$

Then the average velocity estimated from it is:

$v_{average}=\frac{356}{8} \frac{m}{s} =44.5\,\frac{m}{s}$

If the area is approximated instead with the trapezoid form by the green points we used to calculate the acceleration (this would give us an over-estimate):

$Trapezoid\,\, area=(50+42)\,8/2=368$

Then the average velocity estimated from it is:

$v_{average}=\frac{368}{8} \frac{m}{s} =46\,\frac{m}{s}$

while his actual instantaneous velocity seems to be about 46 m/s from the graph