a) Table of Values

t(s) | h(t) (m) | 0 | ht= -4.9(0)2+ 450= 450 | 1 | ht= -4.9(1)2+ 450= 445.1 | 2 | ht= -4.9(2)2+ 450= 430.4 | 3 | ht= -4.9(3)2+ 450= 405.9 | 4 | ht= -4.9(4)2+ 450=371.6 | 5 | ht= -4.9(5)2+ 450=327.5 | 6 | ht= -4.9(6)2+ 450= 273.6 | 7 | ht= -4.9(7)2+ 450= 209.9 | 8 | ht= -4.9(8)2+ 450= 136.4 | 9 | ht= -4.9(9)2+ 450=53.1 | 10 | ht= -4.9(10)2+ 450= -40 |

b) Average velocity for the first 2 seconds after the ball was dropped=

h2-h02-0 = 430.4-4502-0 = -19.62 = -9.8 m/s

c) Average velocity for the following time intervals:

i. 1≤t≤4 = h4-h14-1 = 371.6-445.14-1 = -73.53 = -24.5 m/s

ii. 1≤t≤2 = h2-h12-1 = 430.4-445.12-1 = -14.71 = -14.7 m/s

iii. 1≤t≤1.5 = h1.5-h11.5-1 = h1.5=-4.91.52+450-445.11.5-1 = 438.975-445.11.5-1

= -6.1250.5 = -12.25 m/s = -12.3 m/s

d) Instantaneous velocity at t= 1 second = h1-h0.751-0.75 = 445.1- h0.75= -4.90.752+ 4501-0.75 = 445.1-447.21-0.75 = -2.10.25 = -8.4 m/s

FIND GRAPH ON FOLLOWING PAGE

2. M=10.5-0.4t2, where M is the mass in grams and t is the time in seconds.

a) All the sugar has dissolved when M= 0 g

M=10.5-0.4t2

0=10.5-0.4t2 0.4t2=10.5 t2=10.50.4 t2=26.25 t2=√26.25 t=±5.12 s

Since t is time, the negative value cannot be considered and therefore M is 0 g when t= 5.12 s.

b) Average rate of change in the interval 0≤t≤1 = M1-M01-0 = M1=10.5-0.412-M0=10.5-0.4021-0 = 10.1-10.51-0 = -0.41 = -0.4 g/s

c) Table of Values

t(s) | M(t) (g) | 0 | M=10.5-0.4(0)2= 10.5 | 1 | M=10.5-0.4(1)2= 10.1 |