A cyclist starts from home and rides back and forth along a straight east-west r

Question

A cyclist starts from home and rides back and forth along a straight east-west road. Her velocity is given by the figure below (positive velocities = travel toward the east: negative velocities = travel toward the west). Velocity is measured in m/s and time is measured in minutes. (Plot scale: vertical gridlines are spaced 2 units apart: horizontal gridlines are spaced 5 units apart, so the range of the function is [-9, 9] and its domain is [0, 11].) a. On what time intervals is she stopped? b. How far from home is she the first time she stops, and in what direction? c. At what time does she bike past her house? d. If she maintains her velocity at t = 11, how long will it take her to get back home?

Explanation / Answer

a)
Stopped when v = 0
And this happens when
t = 3 to t = 5
and between t = 9 to 10

So, t = 3 to 5
and t = 9 to 10

-------------------------------------------------------------------

b)
She stops first when t = 3
So, to find the position, we gotta find the area from t = 0 to t = 3
This is a trpzd of base1 = 1 and base 2 = 3
and height = 8
Area= 1/2 * (1 + 3) * 8
= 1/2 * 4 * 8
= 16 * 60(we multiple by 60 to convert minutes to seconds)
= 960

So, she is 960 meters from home.

--------------------------------------------------------------------

c)
The first 3 seconds, she was 960 meters from home.
The negative trpzd area must also be -16

The first trpzd had
b1 = 1 , b2 = 3
When we added bases, we got 4
And height was 8

The second trpzd also has height = 8
So, to negate the area, b1 + b2 = 4
for the second trpzd, this happens when x = 7.5
When x = 7.5, we get b1 = 7.5 - 5 = 2.5
and b2 = 7.5 - 6 = 1.5
And adding 2.5 + 1.5, we get 4

So, at time, t = 7.5 minutes, she rides past her house

-------------------------------------------------------------------

d)
First trapzd area = 16
Second tarpzd area= 1/2(4 + 2) * -8 = -24

Adding them, we get -8

So, we need to find the area of the triangle such that we get +8
so as to return to home base.

That triangle has a line whose slope is
between 10,0 and 11,8, as in, slope = 8

Equation of that line is
y = 8x + C

Using point (10,0), we get C = -80

So, y = 8t - 80

Area of triangle = 1/2 * base * height
WE need this area of triangle to be 8

8 = 1/2 * (t - 10) * (8t - 80)

16 = 8(t - 10)^2

2 = (t - 10)^2

t = 10 + sqrt(2)

Considering time when t= 11,
timw taken to get back home = (10 + sqrt2) - 11
= 10 + sqrt2 - 11
= sqrt2 - 1

So, it will take another sqrt(2) - 1 seconds to make it back home

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.