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maria flor • Dec 26, 2010

Here is the problem:
A man "A" set out from a certain point and traveled at the rate of 6 kph. After "A" had gone for two hours, another man "B" set out to overtake "A" and went 4 km the first hour, 5 km the second hour, 6 km the third hour and so on gaining 1 km every hour. How many hours after "B"left were they together?
Ankita Katdare • Dec 26, 2010
The answer should be 8 Hours.

When B left, A was at 12 km. After 8 hours B met A at a distance of 60 km.

Correct me if I am wrong.
makru921 • Dec 27, 2010
This should make it simple:
After: 1 hour 2hrs 3hrs 4hrs 5hrs 6hrs 7hrs 8hrs 9hrs 10hrs
Distance of A: 6 12 18 24 30 36 42 48 54 60
Distance of B: 0 0 4 9 15 22 30 39 49 60

So the answer is 8 hours after A leaving. Just as AKD says😀
maria flor • Dec 27, 2010
THanks by the way... it was a great help....
maria flor • Dec 27, 2010
Here is another problem:
The velocity of a particle moving along X axis is defined by v=(x^3)-(4x^2) +6x where v is in feet per second and x in feet. Compute the value of the acceleration when x=2ft.
It's a vary simple question....please improve the label.
.
by the way answer is 2ft/(s^2).
diff both side with respect to time and then put the value of x you get the answer.
maria flor • Dec 27, 2010
@ mohit: Your answer is wrong, since you cannot diff both side with respect to time...the variable on the right side is not t it's x.