A car covers a distance of 900 km with a certain speed. If after covering 450 km the speed increases by 15 km, the time required to cover the total distance of 900 km is lessened by 5 hours.What is the original speed of the car in km/hr?
2007-03-11 07:17:48 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Let v be the original speed of the car in km/hr.
Balance by the time,
900/v - [450/v + 450/(v+15)] = 5
Simplify,
v^2+15v - 1350 = 0
Solve for v,
v = 30 km/hr
2007-03-11 07:27:56 · answer #1 · answered by sahsjing 7 · 1⤊ 0⤋
So let's call the original speed of the car x. then the time it would have taken to travel the remaining 450km is 450/x. After increasing the speed by 15 km/h the time it takes is given by 450/(x+15) and we know that is 5 hours less than it would have been if we had not increased our speed. this gives us the equation 450/(x+15) = (450/x) - 5. now just do some simple algebra.
I'll give a brief rundown:
Multiply both sides by x+15 and x
450x = 450(x+15) - 5x(x+15)
distribute
450x = 450x + 6750 - 5x² - 75x
subtract 450x and divide by -5
x² + 15x - 1350 = 0
now factor
(x-30)(x+45) = 0
so our original speed is either 30 Km/h or -45 Km/h
the -45 doesn't make sense so we'll throw it out and our original speed is 30 Km/h
Taa Daa!
2007-03-11 14:39:11 · answer #2 · answered by snilubez 2 · 0⤊ 0⤋
Let the original speed be x km/h.
Then the car covers 900km in 900/x hours.
And it covers 450 km in 450/x hours, obviously.
Then it accelerates and it goes on (x+15) km/h, covering the rest 450 kms in 450/(x+15) hours.
Thus 900/x = 450/x + 450/(x+15) +5, cause the second is 5 hours less.
Now just solve the equation.
450/x = 450/(x+15) + 5
(450(x+15) - 450x)/(x(x+15)) = 5
450*15 = 5* (x^2 + 15x)
1350=x^2 + 15x
x^2 + 15x - 1350 =0
D = 225 + 5400 = 5625 = 75^2
x = (-15 +/- 75)/2.
Thus x = (-15+75)/2 = 30.
Check it:
900kms at 30 km/h is 30 hours.
450 kms at 30 km/h is 15 hours, 450 kms at 30+15=45km/h is 10 hours. 15+10=25, 5 less than 30.
The answer is 30 km/h.
2007-03-11 14:34:00 · answer #3 · answered by --sv-- 2 · 0⤊ 0⤋
Original speed = 30kph.
The original time was 900/30 hours = 30 hours. In the alternative scenario, the first 450km took half of that (15 hours), and the remaining 450km took 450/45 hours = 10 hours, hence cutting 5 hours off the total time.
This result is obtained by the following argument. Initially,
900 / v = t. ......(1)
Subsequently,
(450/v) + 450 / (v + 15) = t - 5 = (900 / v) - 5, by eqn (1).
Simplifying (dividing by 5) and rearranging terms, this leads to the quadratic equation
v^2 + 15 v - 1350 = 0, or (v - 30)(v + 45) = 0.
Only the solution v = 30kph is physical.
Live long and prosper.
2007-03-11 14:27:45 · answer #4 · answered by Dr Spock 6 · 0⤊ 0⤋
Time taken in second instance is five minutes less than in the forst instance.
Thus, t1 - t2 = 5.
t = distance/speed
t1 = 900/x, where x = original speed
t2 = 450/x + 450/(x+15)
900/x - 450/x - 450/(x+15) = 5
450(1/x - 1/(x+15)) = 5
Multiplying by x*(x+15), we get
x^2 +15x - 1350 = 0
Solving, (x - 30)(x + 45) = 0.
Therefore x = 30 or -45.
Since, speed can not be negative,
x i.e. Original speed = 30 km/hr
2007-03-11 14:20:53 · answer #5 · answered by Nimish A 3 · 0⤊ 1⤋
original speed of the car is 30km/h
2007-03-11 14:35:36 · answer #6 · answered by zhihao_chua 1 · 0⤊ 0⤋