Asked by Anonymous
A tourist has planned a trip to cover the distance of 640 miles, driving at some constant speed. However, when he already covered a quarter of the distance, he took a rest for 1.2 hours. Then, in order to arrive at the destination on time, he increased the speed by 20 mph. How long, actually, the trip lasted?
Answers
Answered by
Bosnian
A quarter of the distance = 640 / 4 = 160 miles
rest of the distance = 640 - 160 = 480 miles
Distance, speed, time formula:
s = d / t
If he did not make a rest he would travel:
t = 640 / s1 [ hours ]
where s1 = speed in a first quarter of the distance
t1 = d / s1 = 160 / s1
where t1 =time in a first quarter of the distance
time of a rest = 1.2 [ hours ]
New speed:
s2 = s1 + 20
New time = t2 = rest of the distance / new speed
t2 = 480 / ( s1 + 20 )
In order to arrive at the destination on time mean a trip will last:
640 / s1
how long it would take to travel at a constant speed to lasted s1
Total time:
t = 160 / s1 + 1.2 + 480 / ( s1 + 20 ) = 640 / s1
[ 160 ∙ ( s1 + 20 ) + 1.2 ∙ s1 ∙ ( s1 + 20 ) + 480 ∙ s1 ] / [ s1 ∙ ( s1 + 20 ) ] = 640 / s1
( 160 s1 + 3200 + 1.2 s1² + 24 s1+ 480 s1 ) / [ s1 ∙ ( s1 + 20 ) ] = 640 ∙ ( s1 + 20 ) / [ s1 ∙ ( s1 + 20 ) ]
( 1.2 s1² + 664 s1+ 3200 ) / [ s1 ∙ ( s1 + 20 ) ] = 640 ∙ ( s1 + 20 ) / [ s1 ∙ ( s1 + 20 ) ]
Multiply both sides by s1 ∙ ( s1 + 20 )
1.2 s1² + 664 s1 + 3200 = 640 ∙ ( s1 + 20 )
1.2 s1² + 664 s1 + 3200 = 640 s1 + 12800
Subtract 640 s1 + 12800 to both sides
1.2 s1² + 664 s1 + 3200 - 640 s1 - 12800 = 0
1.2 s1² + 24 s1 - 9600 = 0
Divide both sides by 1.2
s1² + 20 s1 - 8000 = 0
The solutions are:
s1 = - 100
and
s1 = 80
Speed can't be negative so:
s1 = 80 miles / hour
Total time:
t = 160 / s1 + 1.2 + 480 / ( s1 + 20 )
t = 160 / 80 + 1.2 + 480 / ( 80 + 20 )
t = 2 + 1.2 + 480 / 100
t = 2 + 1.2 + 4.8
t = 8 hours
rest of the distance = 640 - 160 = 480 miles
Distance, speed, time formula:
s = d / t
If he did not make a rest he would travel:
t = 640 / s1 [ hours ]
where s1 = speed in a first quarter of the distance
t1 = d / s1 = 160 / s1
where t1 =time in a first quarter of the distance
time of a rest = 1.2 [ hours ]
New speed:
s2 = s1 + 20
New time = t2 = rest of the distance / new speed
t2 = 480 / ( s1 + 20 )
In order to arrive at the destination on time mean a trip will last:
640 / s1
how long it would take to travel at a constant speed to lasted s1
Total time:
t = 160 / s1 + 1.2 + 480 / ( s1 + 20 ) = 640 / s1
[ 160 ∙ ( s1 + 20 ) + 1.2 ∙ s1 ∙ ( s1 + 20 ) + 480 ∙ s1 ] / [ s1 ∙ ( s1 + 20 ) ] = 640 / s1
( 160 s1 + 3200 + 1.2 s1² + 24 s1+ 480 s1 ) / [ s1 ∙ ( s1 + 20 ) ] = 640 ∙ ( s1 + 20 ) / [ s1 ∙ ( s1 + 20 ) ]
( 1.2 s1² + 664 s1+ 3200 ) / [ s1 ∙ ( s1 + 20 ) ] = 640 ∙ ( s1 + 20 ) / [ s1 ∙ ( s1 + 20 ) ]
Multiply both sides by s1 ∙ ( s1 + 20 )
1.2 s1² + 664 s1 + 3200 = 640 ∙ ( s1 + 20 )
1.2 s1² + 664 s1 + 3200 = 640 s1 + 12800
Subtract 640 s1 + 12800 to both sides
1.2 s1² + 664 s1 + 3200 - 640 s1 - 12800 = 0
1.2 s1² + 24 s1 - 9600 = 0
Divide both sides by 1.2
s1² + 20 s1 - 8000 = 0
The solutions are:
s1 = - 100
and
s1 = 80
Speed can't be negative so:
s1 = 80 miles / hour
Total time:
t = 160 / s1 + 1.2 + 480 / ( s1 + 20 )
t = 160 / 80 + 1.2 + 480 / ( 80 + 20 )
t = 2 + 1.2 + 480 / 100
t = 2 + 1.2 + 4.8
t = 8 hours