1.2.1.1
Let s= distance in meters, v= velocity in meters per minute, t= time in minutes.
Distance Time Graph of an object moving along line with a uniform speed
1.2.1.2. The equation of the uniform line motion
s=vt Solve for t
t=vs Given
s=1.8 km=1800 m,v=75 m/min Then
t=75 m/min1800 m=24minA boy takes 24 minutes to walk from home to school.
1.3.1. Let v1= the rate of cleaning a paving by Mpho, m2/h .
Given v1=4 m2/h. The area A of the rectangular paving with the length l and the width w is
A=l×w Given l=12 m,w=6 m. Then
A=12 m×6 m=72 m2 If the rate of cleaning a paving is constant, then
t=vA
Substitute
tI=v1A=4 m2/h72 m2=18 h It will take her18 hours to clean a paving.
1.3.2.Let now v2= the rate of cleaning a paving by Maria, m2/h .
Given v2=4 m2/h=v1. The area A of the rectangular paving with the length l and the width w is
A=l×w Given l=12 m,w=6 m. Then
A=12 m×6 m=72 m2 If the rate of cleaning a paving is constant, then
t=vA
Substitute
tII=v1+v2A=4 m2/h+4 m2/h72 m2=9 h It will take them 9 hours to clean a paving.
1.3.3. Let now v3= the rate of cleaning a paving by Lerato, m2/h .
Given v3=4 m2/h=v1=v2. The area A of the rectangular paving with the length l and the width w is
A=l×w Given l=12 m,w=6 m. Then
A=12 m×6 m=72 m2 If the rate of cleaning a paving is constant, then
t=vA
Substitute
tIII=v1+v2+v3A=4 m2/h+4 m2/h+4 m2/h72 m2=6 h It will take the three of them 6 hours to clean a paving.
1.4.1
R45 for 2.5 kg
The price for washing powder
p1=2.5 kgR45=R18/kg R55.50 for 3 kg
The price for washing powder
p2=3 kgR55.50=R18.50/kg R72 for 4 kg
The price for washing powder
p3=4 kgR72=R18/kg18<18.50
The best price for washing powder is R18/kg.
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