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Chapter 2 Solutions - (calculations)

Problem 1 :

(90 km/hr) * (1/3600) hr/s//N

(0.025 km)/s

Problem 5 (7):
a)

x = 3 t - 4t^2 + t^3 ;

x/.t->1

x/.t2

x/.t3

x/.t4

0

-2

0

12

b) At t = 0 Sec, x = 0 m. At t = 4 Sec, x = 12 m. Then Δx = x_2 - x_1 = 12  - 0 = 12.

c) Between 2 Sec and 4 Sec, x changes from -2 m to 12 m. Then Δx = 12 - (-2) and Δt = 4 - 2, so v_Ave =Δx/Δt= 14/2= 7 M/Sec.
d)

x = 3 t - 4 t^2 + t^3 ;

Plot[x, {t, 0, 4}]

[Graphics:htmlfiles/ch2x_20.gif]

⁃Graphics⁃

Problem 10 (8): See homework solutions

Problem 61: See homework solutions

Problem 96 (1):

(1/160) Hr/Km * 3600 Sec/Hr * 0.0184 Km

0.414 Sec

Problem 116 (44):
a) Call y = 0 at ground level, positive y upwards. Ball lands at a time where y = 0. (A slightly different approach is taken in the homework solutions)

Clear[a, t]

a = -9.8 ;

y_i = 145 ;

y = 0 ;

Solve[y == 0.5 a t^2 + y_i, t]

{{t -5.43984}, {t5.43984}}

Take the positive value.
b)

t = 5.44 ;

Solve[v -a t, v]

{{v53.312}}

c) Now a = 25*9.8 M/s^2and is directed upward. First find t:

Clear[t]

Solve[0 == 9.8 * 25 t - 53.3, t]

{{t0.217551}}

Now use this to find y:

Clear[y]

Solve[y == 0.5 * 9.8 * 25 * .218^2 - 53.3 * .218, y]

{{y -5.79771}}


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