Mr Willson drives 64.8 km from work at a speed of 48 km/h.Mrs Willson drives 81.2 km from work at a speed of 58 om/h.They both leave work at the same time.
a)Who arrives home first.
b)How many minutes later is it before the second person gets back home
Answers
Explanation:
Secondary School Math 5 points
(The distance from town
A to town B is 180 km. If a cyclist rides
half the distance from A to B with a
specific speed and the other half
distance at a speed 15 km/h more
than the first half. He travelled for 5
hours. Find his speed for the first half)
Ask for details Follow Report by Yunaspogo 4 weeks ago
Answers
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Step-by-step explanation:
Distance between town A & B is 180 km
Let
The cyclist covered first half of the distance at the speed of x km/hr
So
his speed for the next half distance = (x+15) km/hr
Now
Distance for the both the halves = 180/2 = 90 km
&
Speed for first half = x km/hr
Speed for second half = (x+15) km/hr
Also
Time = distance / speed
& given that total time taken by him = 5 hrs
Total time taken =
time taken in first half + time taken in second half
=> 90/x + 90/(x+15) = 5
= {(90x + 1350) + 90x}/x(x+15) = 5
= 90x + 1350 + 90x = 5(x^2 + 15x)
= 180x + 1350 = 5x^2 + 75x
= 5x^2 + 75x - 180x - 1350 = 0
= 5x^2 - 105x - 1350 = 0
= x^2 - 21x - 270 = 0
= x^2 - 30x + 9x - 270 = 0
= x(x - 30) + 9(x - 30) = 0
= (x - 30) (x + 9) = 0
= (x - 30) = 0 & (x + 9) = 0
=> x = 30 & - 9
But speed cannot be negative so
X = 30
So speed in his first half = 30 km/hr
please mark in branlient .....
Answer:
a) v=s/t
t=v/s
first case
t =64.8/48
t=1.35
second case
t=81.2/58
t=1.4
so second will arrive first
b) time after subtracting is 0.05hr which is 3 min
Hope it helps
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