Home
Threads
It's all about...rate

It's all about...rate

maria flor

@maria-flor-8cAkif • Oct 22, 2024

Oct 22, 2024

1.2K

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?

0 6

Replies

Welcome, guest

Join CrazyEngineers to reply, ask questions, and participate in conversations.

CrazyEngineers powered by Jatra Community Platform

Ashraf HZ

@Ash • 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.

0
makru921

@makru921-Am65Ah • Dec 26, 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😀

0
maria flor

@maria-flor-8cAkif • Dec 27, 2010

THanks by the way... it was a great help....

0
maria flor

@maria-flor-8cAkif • 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.

0
Morningdot Hablu

@morningdot-6Xuj4M • Dec 27, 2010

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.

0
maria flor

@maria-flor-8cAkif • 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.

0