Question

A man travels from home to town and back in a motorcycle He travels to home from town at a speed which is 20 kilometer/
hour more than his journey to the town from home. The average speed of his total journey was 48 kilometre/hour [3]
(a) If the distance from home to town is 5 kilometre, find his total journey time.
(b) By taking the speed of his journey from home to town as form a second degree equations
ON​

Answer 1

Step-by-step explanation:

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Answer 2

1.Time taken by man in his total journey=31.2 minutes

2.The second degree equation.

$x^2-28x-480=0$

Step-by-step explanation:

1.Average speed of man=48km/h

Distance between home and town=5 km

Total distance traveled by man in his journey=$5+5=10km$

Average speed =$\frac{total\;distance}{total\;time}$

Using the formula

$48=\frac{10}{t}$

$t=\frac{25}{48}=0.52 h$

$t=0.52\times 60=31.2 minutes$

1 hour=60 minutes

2.Let speed of man when he travel from home to town=x km/h

Speed of man when he travel from town to home=(x+20)km/h

Average speed =$\frac{2ab}{a+b}$

Where a=Speed of man when he travel from home to town

b=Speed of man when he travel town to home

Using the formula

$48=\frac{2x(x+20)}{x+x+20}$

$48=\frac{2x(x+20)}{2x+20}=\frac[2x(x+20)}{2(x+10)}$

$48=\frac{x(x+20)}{x+10}$

$x^2+20x=48x+480$

$x^2+20x-48x-480=0$

$x^2-28x-480=0$

This is required second degree equation.

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