Question 60

RACE 1: Chris runs 100 m in time t; Sandy runs 90 m in time t

v(Chris) = 100/t; v (Sandy) = 90/t

RACE 2: s (Chris) = 110 m; s (Sandy) = 100 m

t_{2}(Chris) = s/v = 110t/100 = 1.10t; t_{2}(Sandy) = s/v = 100t/90 = 1.11t

Chris is still faster.

Question 61

Mach 2.5 = 825 m/s; Mach 1.2 = 396 m/s

v = u + at

396 = 825 + a x 30

a = - 14.3 m s^{-2}

Question 62

s

_{1}= 300 m; s_{2}= 300 m; s_{3}= 300 m; s_{4}= 300 m

(1) v_{1}^{2}= u^{2}+ 2as

= 0 + 2 x 0.95 x 300

= 23.87 m s^{-1}

v = u + at

23.87 = 0 + 0.95 t

t_{1}= 25.1 s

(2) v_{2}= s_{2}/t_{2}

t_{2}= s_{2}/ v_{2}= 300/23.87 = 12.57 s;v_{2}= 23.87 m s^{-1}

(3) t_{3}= 12.57 s; v_{3}= 23.87 m s^{-1}

(4) v^{2}= u^{2}+ 2as

0^{2}= 23.87^{2}+ 2 x a x 300

a = - 0.95 m s^{-2}

t_{4}= t_{1}= 25.1 s

Answers: (a) Maximum speed = 23.87 m s^{-1}

(b) Time = t_{1}+ t_{2}+ t_{3}+ t_{4}= 75.3 s

(c)

Question 63

The four intervals are shown in the following diagram

At top of jump v_{4}= 0

v^{2}= u^{2}+ 2as

0^{2}= u^{2}+ 2 x -10 x +0.76

u (take-off speed (v_{4})) = 3.9 m/s

BOTTOM 15 cm:

v_{2}^{2}= u^{2}+ 2 as_{2}

= 3.9^{2}+ 2 x -10 x 0.15

v_{2}= 3.49 m/s

v_{2}= u + at_{2}

t_{2}= (v_{2}- u)/a = 0.04 s

TOP 15 cm:

v_{3}= u^{2}+ 2as

= 3.9^{2}+ 2 x -10 x 0.61

v_{3}= 1.73 m/s

t_{3}= (v_{3}- u)/a = 0.22 s

t_{4}= (v_{4}- u)/a = 0.39 s

Answers:

(a) Bottom 15 cm: t_{1}to t_{2}= 0.04 s

(b) Top 15 cm: t_{3}to t_{4}= 0.17 s

(c) Therefore, time at bottom is about 1/4 of time at top.

Question 64

First toss: s

_{1}, t_{1}, u_{1}, a

Second toss: s_{2}, 2t_{1}, u_{2}, a

At top:

First toss: v = u_{1}+ at_{1}= 0

u_{1}= - at_{1}

Second toss: v = u_{2}+ at_{2}= 0

u_{2}= - at_{2}which equals -2at_{1}

Hence: u_{1}= -at_{1}= u_{2}/2

s_{1}= u_{1}t_{1}+ ½ at_{1}^{2}

s_{2}= u_{2}t_{2}+ ½ at_{2}^{2}

s_{2}= 2u_{1}2t_{1}+ ½ a(2t_{1})^{2}

s_{2}= 4u_{1}t_{1}+ 2at_{1}^{2}

Ratio s_{2}/s_{1}= (4u_{1}t_{1}+ 2at_{1}^{2})/ (u_{1}t_{1}+ ½ at_{1}^{2})

= (4u_{1}+ 2 at_{1})/(u_{1}+ ½ at_{1})

Replace at_{1}by -u_{1}

Ratio s_{2}/s_{1}= (4u_{1}+ -2 u_{1})/(u_{1}+ ½ u_{1})

= 2u_{1}/½ u_{1}

Therefore: s_{2}is 4 times s_{1}or second toss goes four times as high.

Question 65

(a) s = -50m; t

_{2}= t_{1}- 1; u_{1}= 0; u_{2}= ?

Race 1:

s = ut

_{1}+ ½ at_{1}^{2 }-50 = ½ x^{-}10 t_{1}^{2 }t_{1}= 3.2 s; hence t_{2}= 2.2 s

v_{1}= u + at = 0 +^{-}10 x 3.2 =^{-}32 ms^{-1 }

Race 2:

s = u

_{2}t_{2}+ ½ at_{2}^{2 }-50 = u_{2}x 2.2 + ½ x^{-}10 x 2.2^{2 }u_{2}= -11.7 ms^{-1}; v_{2}= u +at = -11.7 +^{-}10 x 2.2 =^{-}33.7 ms^{-1 }

(b)

Question 66

Acceleration is more thrilling!

a (Kitty) = (v-u)/t = (628 x 1000/3600)/3.72 = 46.9 ms^{-1 }a (Eli) = (v-u)/t = (116 x 1000/3600)/0.04 = 805 ms^{-1 }

**Question 67
**

Case 1: starting velocity =

^{+}u

v

_{1}^{2}= u^{2}+ 2as

v_{1}= Ö (u^{2}+ 2as)Case 2: starting velocity =

^{-}u

v

_{2}^{2}= u^{2}+ 2as

v_{2}= Ö (u^{2}+ 2as)

The formula is the same for both because when u is squared it doesn't matter if the value for u is + or -. Hence, the velocity is the same in both cases.

Question 68

s

_{1}= ut_{1}+ ½ at_{1}^{2 }s_{1}=^{-}5t_{1}^{2 }s_{2}= u(t_{1}- 1) + ½ a(t_{1}- 1)^{2 }s_{2}=^{-}5 (t_{1}- 1)^{2 }s_{2}=^{-}5t_{1}^{2}+ 10t_{1}- 5

(a) s_{1}- s_{2}= 5 - 10t_{1 }(or, alternatively: s_{2}- s_{1}= 10t_{1 }- 5). This is actually the difference in displacement. The difference in "distance" (as the question asks) is the absolute value of the answer. Thank you to Vic Braun for pointing out the difference.

(b) v_{1}= u_{1}+ at

v_{1}= -10t_{1 }v_{2}= u_{2}+ a(t_{1}- 1)

v_{2}= -10t_{1}+ 10

Ratio v_{1}/v_{2}= -10t_{1}/(-10t_{1}+ 10) = t_{1}/(t_{1}- 1)

Question 69

Case 1

s = ut + ½ at

^{2 }400 = 0 + ½ a x 7.2^{2 }a = 15.4 ms^{-2}Case 2

v = u + at

305 x 1000/3600 = 0 + a x 7.2

a = 11.8 ms^{-2 }The rider probably reached top speed before the 400 m so in Case 2 where he reached 305 kmh^{-1}at the finish line, he probably reached this after 5 seconds and kept constant speed from then on. In case 1, we are assuming acceleration is constant and that the top speed is irrelevant. The v/t graphs probably look like this:

Question 70

At current mass: t

_{1}= 6.92s and s = 400 m

s = ut + ½ at

_{1}^{2 }400 = 0 + ½ a x 6.92^{2 }a = 16.7 ms^{-2}At heavier mass: 60 pounds - 60/7.2 = 27.3 kg

Time lost = 20/27.3 x 0.3 s = 0.22 s

t_{2}= 6.92 + 0.22 s = 7.14 s

s = ut +½ at^{2}

400 = 0 + ½ a x 7.14^{2 }a = 15.7 ms^{-2 }

**Question 71
**

(a) 30 knots = 30 x 100 feet in 1 minute = 30 x 100 x 60 ft per hr

= 180000 ft/hr = 180000/6080 nautical miles/hr = 30 naut mi/hr

(b) 30 knots - 180000 ft/hr = 180000 x 0.305 metres/hr = 180 x 0.305 km/h

= 54.9 km h^{-1 }

**Question 72
**

t (s)

s (m) Earth

s (m) Moon

s (m) Mars

s (m) Sun

0.0

0.00

0.00

0.00

0.00

0.2

0.20

0.03

0.08

5.40

0.4

0.80

0.13

0.30

21.60

0.6

1.80

0.29

0.68

48.60

0.8

3.20

0.51

1.22

86.40

1.0

5.00

0.80

1.90

135.00

1.2

7.20

1.15

2.74

194.40

1.4

9.80

1.57

3.72

264.60

1.6

12.80

2.05

4.86

345.60

1.8

16.20

2.59

6.16

437.40

2.0

20.00

3.20

7.60

540.00

**Question 73** - The 30-Year Problem

There are at least five solutions to this problem. Perhaps the easiest (?) one for Senior Physics students is one using internal triangles.

1. Drop perpendiculars from points B and D to line AC (see figure below):

2. Let the constant speed = v (miles/minute). Then: distance (miles) = time (minutes) x speed (v)

The distance AC (directly) = 30v; the distance ADC = 40v; the distance ABC = 35v.

Hence: AD = 40v -10; AB = 35v - 10

Let x be the distance from C to the perpendicular on AC from B.

Let y be the distance from C to the perpendicular on AC from D.

The two lower triangles are similar, hence the perpendiculars have the lengths x and y as shown in the figure.

3. Using Pythagorean theorem on the top two triangles and one of the two identical (similar) lower triangles:

Equation 1: (40v-10)^{2}= x^{2}+ (30v-y)^{2}

Equation 2: (35v-10)^{2}= y^{2}+ (30v-x)^{2}

Equation 3: 10^{2}= x^{2}+ y^{2}

4. Expand Equation 1 and substitute 10^{2}- y^{2}for x^{2}

1600v^{2}- 800 v + 100 = 100 - y^{2}+ 900v^{2}- 60 vy + y^{2}

700v^{2}- 800 v = -60vy

y = 13.33 - 11.67v

5. Repeat for Equation 2 substituting 10^{2}- x^{2}for y^{2}:

1225v^{2}- 700v +100 = 100 - x^{2}+ 900v^{2}- 60vx^{2}

325v - 700v = -60vx

x = 11.67 - 5.42v

6. Substitute these values for x and y into equation 3:

100 = (11.67 - 5.42v)^{2}+ (13.33 - 11.67v)^{2}

100 = 136.1 - 126.4v + 29.38v^{2}+ 177.8 - 311.1v + 136.1v

165.5v^{2}- 437.5v + 213.9 = 0

7. Use quadratic formula to solve for v

Two solutions: v = 0.648 miles/minute and v = 1.996 miles/minute.

8. Convert to miles/hour (x 60): 38.8 mph and 119.8 mph.

9. Discard the second speed as too fast (break the speed limit?).

Answer: 38.8 mph

ALTERNATIVE SOLUTION:

Use the cos rule: the cosine of any angle of a triangle equals the sum of the squares of the adjacent sides less the square of the other side, all divided by twice the product of the adjacent sides.

Applying this to the figure (and simplifying) shows that:

Cosine of angle ABC = (7 - 3.25v)/6

Cosine of angle ACD = (8-7v)/6

When these two values are squared and added together they equal 1.

This follows from the rule that: when two angles add up to 90°, the sum of squares of their cosines = 1.

Simplifying and collecting terms gives a quadratic which produces two roots as in the solution above.

ANOTHER SOLUTION

Construct a rectangle as shown in the figure below:

1. Write square-of-the-hypotenuse equations for the three right triangles of which AC, AD and AB are the hypotenuses.

Equation 1 (for AC): (30v)^{2}= (10 + x)^{2}+ (y + 10)^{2}

Equation 2: (for AD) (40v-10)^{2}= x^{2}+ (y + 10)^{2}

Equation 3: (for AB): (35v - 10)^{2}= y^{2}+ (x + 10)^{2}

2. Subtract Equation 2 from Equation1 and find x in terms of v.

30v^{2}- (40v - 10)^{2}= (x + 10)^{2}- x^{2}

x = -35v^{2}+ 40v - 10

3. Subtract Equation 3 from Equation 1 and find y in terms of v.

30v^{2}- (35v - 10)^{2}= (y + 10)^{2}- y^{2}

y = -16.25v^{2}+ 35v - 10

4. Substitute these values of x and y into Equation 1.

(30v)^{2}= (-16.25v^{2}+ 35v)^{2}+ (-35v^{2}+ 40v)^{2}

900v^{2}= 16.25^{2}v^{4}- 16.25 x 35 x 2 v^{3}+ 35^{2}v^{2}+ 35^{2}v^{4}- 35 x 40 x 2 v^{3}+ 40^{2}v^{2}

5. Simplify and divide by v^{2}, (allowable since v can't be equal to zero).

1489.1v^{2}- 3937.5v + 1925 = 0

6. You should have a quadratic. Solve using the quadratic formula and you should get roots as in the methods above.

Pretty simple huh?

Question 74

First statement: s µ t

^{2}is correct as our formula is s = ½ at^{2}which is the same as s µ t^{2 }Second statement: v µ s is incorrect as v^{2}= 2as or v^{2}µ s

**Question 75
**

tan 4.3° = 1000/s therefore s = 1000/tan 4.3° = 13300 m

298 km/h = 298 x 1000/3600 ms^{-1}= 82.8 ms^{-1}

t = s/v = 13300/82.8 = 160 s

NOTE: the answer in the back of the book is wrong. It is not 44.6 s but 160 s.

**Question 76
**

Impact speeds:

Of a child in a car: v = 50 x 1000/3600 = 13.9 ms^{-1 }Falling from a building: (assume each story to be about 3 m high):

v^{2}= u^{2}+ 2as

v = Ö (2 x 10 x 6) = 10.9 ms^{-1 }Hence there is a lower impact speed in falling from a building but the concrete is harder than a car dashboard. Probably about equal impact damage.

**Question 77**

s = ut + ½ at

^{2 }64 = 0 + ½ x 2 x t^{2 }t = 8.0 s

s_{(t=0) }= 0

s_{(t=2)}= ½ at^{2}= ½ a 2^{2}= 4 m

s_{(t=4) }= ½ a x 4^{2}= 16 m

s_{(t=6) }= ½ a x 6^{2}= 36 m

s_{(t=8)}= ½ a x 8^{2}= 64 m