Question

An earthworm of length 15cm is slivering along at 2cm/s. A beetle overtakes the worm in 5 seconds. How fast was the beetle moving???

Answer 1

The point at which the beetle overtakes the worm is when his position is equal to the farthest position of the worm. Assuming the timer started when the beetle was at the beginning point of the worm, the total distance traveled by the beetle must equal the length of the worm, plus any distance covered by the worm in that time:
\[v_{b}t=L+v_{w}t\]\[v_{b}=\frac{L+v_{w}t}{t}\]\[v_{b}=\frac{15cm+(2cm/s)(5s)}{5s}=\frac{25cm}{5s}=5\frac{cm}{s}\]

Answer 2

what formula is that?

Answer 3

It's not a standard formula, other than distance traveled being velocity times time. It's just thinking about how you would model the given problem mathematically.

Answer 4

why do you multiply 2 cm/s by 5s?

Answer 5

to get the distance

Answer 6

It said it took 5 seconds for the beetle to pass the worm. The worm is travelling at 2cm/s, so in 5 seconds, the worm travelled (2x5)=10 cm. You have to take this into account because the worm is not standing still while the beetle is walking by; the beetle has to cover any distance travelled by the worm while the beetle is travelling.

Answer 7

Thanks man