Recitation Week 3 Chapter 2 Problem 2.5. A position-time graph for a particle moving along the x axis is shown in Figure P2.5. (a) Find the average velocity in the time interval t = 1.50 s to t = 4.00 s. (b) Determine the instantaneous velocity at t = 2.00 s by measuring the slope of the tangent to the graph. (c) At what value of t is the velocity zero?

xm

12 9 6 3 0

0 1 2 3 4 5 6 ts

(a) The average velocity is the total displacement over the elapsed time, so vavg =

x(4.00 s) − x(1.50 s) 2.0 m − 8.0 m = = −2.4 m/s 4.00 s − 1.50 s 2.5 s

(1)

(b) Using rise-over-run to determine the tangent slope v(t = 2.00 s) =

0 m − 11.5 m = −3.8 m/s 3.5 s − 0.5 s

(2)

(c) Looking for the minimum of x(t) (where the tangent curve is flat), we see that v(4.0 s) = 0. Problem 2.11. A hare and a tortoise compete in a race over a straight course 1.00 km long. The tortoise crawls at a speed of 0.200 m/s toward the finish line. The hare runs at a speed of 8.00 m/s toward the finish line for 0.800 km and then stops to tease the slow-moving tortoise as the tortoise eventually passes by. The hare waits for a while after the tortoise passes by and then runs toward the finish line again at 8.00 m/s. Both the hare and the tortise cross the finish line at exactly the same instant. Assume both animals, when moving, move steadily at their respective speeds. (a) How far is the tortoise from the finish line when the hare resumes the race? (b) For how long in time was the hare stationary? Sometimes it is useful to draw a graph to get a feel for what’s going on.

t

hare

hare

pause

e

ois

t tor

x (a) The final distance run by the hare is xh,2 = L − ∆xh,1 which takes the hare th,2 =

L − xh,1 vh

(3) (4)

In this time, the tortoise covers xt,2 = vt · th,2 = vt =

vt L − xh,1 = (L − xh,1 ) vh vh

0.200 m/s (1.00 km − 0.800 km) = 5.00 m 8.00 m/s

(5) (6)

(b) The hare is running for th,run =

L vh

(7)

tt,run =

L vt

(8)

The tortoise is running for

and the tortoise is running for the whole race, so the hare pauses for L 1.00 km 1.00 km L − = − vt vh 0.200 m/s 8.00 m/s = 4.88 ks = 1.35 hours

th,pause = tt,run − th,run =

(9) (10)

Problem 2.29. The driver of a car slams on the brakes when he sees a tree blocking the road. The car slows uniformly with an acceleration of −5.60 m/s2 for 4.20 s, making straight skid marks 62.4 m long, all the way to the tree. With what speed does the car then strike the tree? The speed of the car is given by v(t) = at + v0

(11)

and its position is given by x(t) =

1 2 at + v0 t + x0 2

(12)

We know a, t, and x − x0 , so we can use the second equation to find v0 . x − x0 − 12 at2 t

v0 =

(13)

We can plug this into the first equation to find the final velocity. v = at +

x − x0 − 12 at2 t

= −5.60 m/s2 · 4.20 s +

(14) 62.4 m −

1 2

· (−5.60 m/s2 ) · (4.20 s)2 4.20 s

= 3.10 m/s

(15) (16)

Problem 2.33. An object moves with constant acceleration 4.00 m/s2 and over a time interval reaches a final velocity of 12.0 m/s. (a) If its initial velocity is 6.00 m/s, what is its displacement during the time interval? (b) What is the distance it travels during the time interval? (c) If its initial velocity is −6.00 m/s, what is its displacement during the time interval? (d) What is the total distance it travels during the interval in (c)? (a) The position of an object undergoing constant acceleration is x(t) =

1 2 at + v0 t + x0 2

Solving for time using the quadratic formula p p −v0 ± v02 − 4(a/2) · (x0 − x) −v0 ± v02 − 2a(x0 − x) = t= 2(a/2) a

(17)

(18)

Plugging this into the velocity formula p

v02 − 2a(x0 − x) v = at + v0 = a · + v0 a q q = −v0 ± v02 − 2a(x0 − x) + v0 = ± v02 + 2a(x − x0 ) −v0 ±

The ±

√

(19) (20)

is annoying, so square both sides v 2 = v02 + 2a(x − x0 )

(21)

Solving for displacement x − x0 =

v 2 − v02 12.02 − 6.002 = m = 13.5 m 2a 2 · 4.00

(22)

(b) With positive accerleration and a positive initial velocity, the objects velocity will always be increasingly positive, so the distance traveled is the same as the displacement. Db = 13.5 m (23) (c) A negative initial velocity has no effect on the answer to (a), because v only shows up as a square: x − x0 =

v 2 − v02 12.02 − (−6.00)2 = m = 13.5 m 2a 2 · 4.00

(24)

(d) With positive accerleration and a negative initial velocity, the objects velocity will drop to zero, after which it will be increasingly positive. The total distance traveled is therefore the distance traveled in the negative direction while the velocity is decreasingly negative, plus the distance traveled in the positive direction while velocity is increasingly positive. 0 − v02 v 2 − 0 v 2 + v02 12.02 + (−6.00)2 = + = m = 22.5 m (25) Dd = 2a 2a 2|a| 2 · 4.00

xm

Problem 2.61. Kathy tests her new sports car by racing with Stan, an experienced racer. Both start from rest, but Kathy leaves the starting line 1.00 s after Stan does. Stan moves with a constant acceleration of 3.50 m/s2 , while Kathy maintains an acceleration of 4.90 m/s2 . Find (a) the time at which Kathy overtakes Stan, (b) the distance she travels before she catches him, and (c) the speeds of both cars at the instant Kathy overtakes Stan.

80 60 40 20 0

0

1

2

3

4

5

6

7 ts

(a) Kathy overtakes Stan when their positions match. xk =

1 1 ak (tp − ∆t )2 = xs = as t2p 2 2 ak (tp − ∆t )2 = as t2 √ √ ak (tp − ∆t ) = ± as tp √ √ √ tp ( ak ∓ as ) = ak ∆t p √ ak ∆t 4.90 m/s2 · 1.00 s p tp = √ = 6.46 or 0.542 s √ =p ak ∓ as 4.90 m/s2 ∓ 3.50 m/s2

(26) (27) (28) (29) (30)

The smaller time occurs before Kathy starts, where the quadratic equation for xk (t) does not hold (xk (t < ∆(t)) = 0, as Kathy is sitting at the starting line). Therefore, the Kathy passes Stan at the larger time tp = 6.46 s (measured since Stan left the starting line). (b) Kathy travels the same distance as Stan xs (tp ) =

1 2 1 as t = (3.50 m/s2 ) · (6.46 s)2 = 73.0 m 2 p 2

(31)

Symbolically, 1 xs (tp ) = as 2

√

ak ∆t √ √ ak ∓ as

2 =

as ak ∆2t √ √ 2( ak ∓ as )2

(32)

(c) The velocities at the passing point are vs (tp ) = as tp = (3.50 m/s2 ) · (6.46 s) = 22.6 m/s 2

vk (tp ) = ak (tp − ∆t ) = (4.90 m/s ) · (6.46 s − 1.00 s) = 26.7 m/s

(33) (34)

xm

12 9 6 3 0

0 1 2 3 4 5 6 ts

(a) The average velocity is the total displacement over the elapsed time, so vavg =

x(4.00 s) − x(1.50 s) 2.0 m − 8.0 m = = −2.4 m/s 4.00 s − 1.50 s 2.5 s

(1)

(b) Using rise-over-run to determine the tangent slope v(t = 2.00 s) =

0 m − 11.5 m = −3.8 m/s 3.5 s − 0.5 s

(2)

(c) Looking for the minimum of x(t) (where the tangent curve is flat), we see that v(4.0 s) = 0. Problem 2.11. A hare and a tortoise compete in a race over a straight course 1.00 km long. The tortoise crawls at a speed of 0.200 m/s toward the finish line. The hare runs at a speed of 8.00 m/s toward the finish line for 0.800 km and then stops to tease the slow-moving tortoise as the tortoise eventually passes by. The hare waits for a while after the tortoise passes by and then runs toward the finish line again at 8.00 m/s. Both the hare and the tortise cross the finish line at exactly the same instant. Assume both animals, when moving, move steadily at their respective speeds. (a) How far is the tortoise from the finish line when the hare resumes the race? (b) For how long in time was the hare stationary? Sometimes it is useful to draw a graph to get a feel for what’s going on.

t

hare

hare

pause

e

ois

t tor

x (a) The final distance run by the hare is xh,2 = L − ∆xh,1 which takes the hare th,2 =

L − xh,1 vh

(3) (4)

In this time, the tortoise covers xt,2 = vt · th,2 = vt =

vt L − xh,1 = (L − xh,1 ) vh vh

0.200 m/s (1.00 km − 0.800 km) = 5.00 m 8.00 m/s

(5) (6)

(b) The hare is running for th,run =

L vh

(7)

tt,run =

L vt

(8)

The tortoise is running for

and the tortoise is running for the whole race, so the hare pauses for L 1.00 km 1.00 km L − = − vt vh 0.200 m/s 8.00 m/s = 4.88 ks = 1.35 hours

th,pause = tt,run − th,run =

(9) (10)

Problem 2.29. The driver of a car slams on the brakes when he sees a tree blocking the road. The car slows uniformly with an acceleration of −5.60 m/s2 for 4.20 s, making straight skid marks 62.4 m long, all the way to the tree. With what speed does the car then strike the tree? The speed of the car is given by v(t) = at + v0

(11)

and its position is given by x(t) =

1 2 at + v0 t + x0 2

(12)

We know a, t, and x − x0 , so we can use the second equation to find v0 . x − x0 − 12 at2 t

v0 =

(13)

We can plug this into the first equation to find the final velocity. v = at +

x − x0 − 12 at2 t

= −5.60 m/s2 · 4.20 s +

(14) 62.4 m −

1 2

· (−5.60 m/s2 ) · (4.20 s)2 4.20 s

= 3.10 m/s

(15) (16)

Problem 2.33. An object moves with constant acceleration 4.00 m/s2 and over a time interval reaches a final velocity of 12.0 m/s. (a) If its initial velocity is 6.00 m/s, what is its displacement during the time interval? (b) What is the distance it travels during the time interval? (c) If its initial velocity is −6.00 m/s, what is its displacement during the time interval? (d) What is the total distance it travels during the interval in (c)? (a) The position of an object undergoing constant acceleration is x(t) =

1 2 at + v0 t + x0 2

Solving for time using the quadratic formula p p −v0 ± v02 − 4(a/2) · (x0 − x) −v0 ± v02 − 2a(x0 − x) = t= 2(a/2) a

(17)

(18)

Plugging this into the velocity formula p

v02 − 2a(x0 − x) v = at + v0 = a · + v0 a q q = −v0 ± v02 − 2a(x0 − x) + v0 = ± v02 + 2a(x − x0 ) −v0 ±

The ±

√

(19) (20)

is annoying, so square both sides v 2 = v02 + 2a(x − x0 )

(21)

Solving for displacement x − x0 =

v 2 − v02 12.02 − 6.002 = m = 13.5 m 2a 2 · 4.00

(22)

(b) With positive accerleration and a positive initial velocity, the objects velocity will always be increasingly positive, so the distance traveled is the same as the displacement. Db = 13.5 m (23) (c) A negative initial velocity has no effect on the answer to (a), because v only shows up as a square: x − x0 =

v 2 − v02 12.02 − (−6.00)2 = m = 13.5 m 2a 2 · 4.00

(24)

(d) With positive accerleration and a negative initial velocity, the objects velocity will drop to zero, after which it will be increasingly positive. The total distance traveled is therefore the distance traveled in the negative direction while the velocity is decreasingly negative, plus the distance traveled in the positive direction while velocity is increasingly positive. 0 − v02 v 2 − 0 v 2 + v02 12.02 + (−6.00)2 = + = m = 22.5 m (25) Dd = 2a 2a 2|a| 2 · 4.00

xm

Problem 2.61. Kathy tests her new sports car by racing with Stan, an experienced racer. Both start from rest, but Kathy leaves the starting line 1.00 s after Stan does. Stan moves with a constant acceleration of 3.50 m/s2 , while Kathy maintains an acceleration of 4.90 m/s2 . Find (a) the time at which Kathy overtakes Stan, (b) the distance she travels before she catches him, and (c) the speeds of both cars at the instant Kathy overtakes Stan.

80 60 40 20 0

0

1

2

3

4

5

6

7 ts

(a) Kathy overtakes Stan when their positions match. xk =

1 1 ak (tp − ∆t )2 = xs = as t2p 2 2 ak (tp − ∆t )2 = as t2 √ √ ak (tp − ∆t ) = ± as tp √ √ √ tp ( ak ∓ as ) = ak ∆t p √ ak ∆t 4.90 m/s2 · 1.00 s p tp = √ = 6.46 or 0.542 s √ =p ak ∓ as 4.90 m/s2 ∓ 3.50 m/s2

(26) (27) (28) (29) (30)

The smaller time occurs before Kathy starts, where the quadratic equation for xk (t) does not hold (xk (t < ∆(t)) = 0, as Kathy is sitting at the starting line). Therefore, the Kathy passes Stan at the larger time tp = 6.46 s (measured since Stan left the starting line). (b) Kathy travels the same distance as Stan xs (tp ) =

1 2 1 as t = (3.50 m/s2 ) · (6.46 s)2 = 73.0 m 2 p 2

(31)

Symbolically, 1 xs (tp ) = as 2

√

ak ∆t √ √ ak ∓ as

2 =

as ak ∆2t √ √ 2( ak ∓ as )2

(32)

(c) The velocities at the passing point are vs (tp ) = as tp = (3.50 m/s2 ) · (6.46 s) = 22.6 m/s 2

vk (tp ) = ak (tp − ∆t ) = (4.90 m/s ) · (6.46 s − 1.00 s) = 26.7 m/s

(33) (34)