how far behind is the other one?
2007-09-13 12:51:37 · 4 answers · asked by abbs 2 in Science & Mathematics ➔ Mathematics
First Ant Time: 71 / 4.98 = 14.257
Second Ant Time: 71 / 3 = 23.667
Difference Between Times: 23.667 - 14.257 = 9.41 Seconds
That's how far the second ant is behind in time.
To find the distance simply multiply by how much distance the ant can cover each second in this case 3cm/s.
9.41 * 3 = 28.23cm
So the Answer:
9.41 Seconds
or
28.23 cm
2007-09-13 13:05:32 · answer #1 · answered by AndrewB 3 · 0⤊ 0⤋
fast ant covers table in
71/4.98 sec
slow ant cmpletes
3 * 71/4.98 cm in 3 sec.
distance remaining for second ant to finish = how far behind second ant is when the first one completes 71 cm so:
71 - (3*71/4.98) = 28.22cm
2007-09-13 20:03:10 · answer #2 · answered by 037 G 6 · 0⤊ 0⤋
Let t_1 represent the fast ant
Let t_2 represent the slow ant
t_1 = d/v
= 71 / 4.98
=14.257 s
t_2 = d/v
=71 / 3
=23.6 s
Difference of time 23.6 - 14.257 = 9.4s
v=d/t
d=vt
=3*9.4
=23.2 cm
2007-09-13 20:04:28 · answer #3 · answered by de4th 4 · 0⤊ 0⤋
how can a ant travel 4.98cms a second
2007-09-13 19:56:22 · answer #4 · answered by Anonymous · 0⤊ 3⤋