2006-08-19 22:46:54 · 8 answers · asked by sudhish n 1 in Science & Mathematics ➔ Mathematics
66.8 kmph, well that is what you guessed:)
2006-08-19 22:54:59 · answer #1 · answered by G♥♥G♥♥ღ 4 · 0⤊ 0⤋
Car should leave 2 minutes earlier. Drive at 60 kmph. Always 2 minutes ahead.
If the car drives less than or more than 60 kmph, the car may be 2 minutes ahead for an instant. Otherwise the truck will gain or fall further back.
2006-08-19 23:09:28 · answer #2 · answered by SPLATT 7 · 0⤊ 0⤋
The question is not very clear.
If you mean the car is to travel the same distance and arrive there 2 minutes early then the truck, then these are the calculations:
S - Speed
D - Distance
T - Time
Speed = Distance / time.
S = D/T = (60Km)/1hr. = 60Km/h. (for truck).
For 2 minutes less:
58 / 60 = 0∙96666.....hr. (that is, traveling time of car in hours).
S = D/T
S = (60Km) / (0∙96666....hr) = 62∙06896552...Km/h.
S = 62∙06896552...Km/h. (Speed of car).
2006-08-20 03:57:19 · answer #3 · answered by Brenmore 5 · 0⤊ 0⤋
62 kms/hr.
Data in question are not sufficient.A car can travel at 2 minutes ahead of the truck at any speed with given data.
2006-08-20 00:03:59 · answer #4 · answered by Anonymous · 0⤊ 0⤋
This question can't be answered without a distance. Either you wrote the question wrong or the answer is: more information is needed.
2006-08-20 00:03:18 · answer #5 · answered by Elim 5 · 0⤊ 0⤋
Question is wrong. Data is insufficient. It all depends on the distance between them.
2006-08-19 23:18:22 · answer #6 · answered by steelrooter 2 · 0⤊ 0⤋
Not enough info
2006-08-20 01:02:14 · answer #7 · answered by MollyMAM 6 · 0⤊ 0⤋
question incomprehensible
2006-08-19 22:54:38 · answer #8 · answered by raj 7 · 0⤊ 0⤋