Two cars left the city for a suburb, 120 km away, at the same time. The speed of one of the cars was 20 km/hour greater than the speed of the other, and that is why it arrived at the suburb 1 hour before the other car. What is the speed of both cars.?

Answer 1

Slower car: #40"km/h"#
Faster car: #60"km/h"#

Let #x=#hours it took the slower car to reach the suburb The time it took the faster car to reach the suburb is #x-1#
Since distance#div#time#=#speed, The speed of the slower car is #120"km"div(x)"h"=120/x"km/h"# The speed of the faster car is #120"km"div(x+1)"h"=120/(x-1)"km/h"#
Since the faster car travelled #20"km/h"# faster than the slower car, #120/(x-1)=120/x+20#
Now multiply both sides by #(x-1)# and then #x# to start solving. #(120(x-1))/(x-1)=(120(x-1))/x+20(x-1)# #(120x(x-1))/(x-1)=(120x(x-1))/x+20x(x-1)#
Simplify, distribute, factor, disregard, etc. Essentially, solve. #(120xcolor(red)cancelcolor(black)(x-1))/color(red)cancelcolor(black)(x-1)=(120color(red)cancelcolor(black)x(x-1))/color(red)cancelcolor(black)x+20x(x-1)# #color(red)cancelcolor(black)(120x)=color(red)cancelcolor(black)(120x)-120+20x^2-20x# #0=20x^2-20x-120# #0=x^2-x-6# #0=x^2+2x-3x-6# #0=x(x+2)-3(x+2)# #0=(x+2)(x-3)# #x+2=0orx-3=0# #x="-"2orx=3#
Since #x# is the number of hours for the slower car to reach the suburb, a negative solution can be disregarded as extraneous as negative time does not (yet?) exist. #x=3#
We know that the speed of the slower car is #120/x"km/h"# and that of the faster car is #120/(x-1)"km/h"# so plugging in #x#, we get: Slower car: #120/((3))=120/3=40"km/h"# Faster car: #120/((3)-1)=120/2=60"km/h"#
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

Let the speed of the slower car be ( x ) km/hour and the speed of the faster car be ( x + 20 ) km/hour.

Using the formula: ( \text{speed} = \frac{\text{distance}}{\text{time}} )

For the slower car: ( \frac{120}{x} ) hours

For the faster car: ( \frac{120}{x+20} ) hours

Given that the faster car arrives 1 hour earlier:

( \frac{120}{x} = \frac{120}{x+20} + 1 )

Solving for ( x ): ( 120(x+20) = 120x + x(x+20) )

( 120x + 2400 = 120x + x^2 + 20x )

( x^2 + 20x - 2400 = 0 )

( (x - 40)(x + 60) = 0 )

( x = 40 ) or ( x = -60 )

Since speed cannot be negative, the speed of the slower car is 40 km/hour and the speed of the faster car is ( 40 + 20 = 60 ) km/hour.

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7