# What constant acceleration do you need to catch up?

1. Mar 14, 2017

### Cynthia

1. The problem statement, all variables and given/known data
Large cockroaches can run as fast as 1.50 m/s in short bursts. Suppose you turn on the light in a cheap motel and see one scurrying directly away from you at a constant 1.50 m/s. If you start 0.91 m behind the cockroach with an initial speed of 0.82 m/s toward it, what minimum constant acceleration would you need to catch up with it when it travelled 1.20 m, just short of safety under a counter?

2. Relevant equations
x=x0+v0*t+1/2*a*t^2

3. The attempt at a solution
I thought that to solve this I musty first find the time it took the cockroach to get to the counter fro 0.91m to 1.2 m. To do this I used the formula x=x0+v0*t+1/2*a*t^2 (where x=1.2, x0=0.91, v0=1.5m/s and, a=0 and t=?)

so I got t= (x-x0)/v0
t= 0.29/1.5 =0.193 seconds

Then I used x=x0+v0*t+1/2*a*t again and rearranged it to find the acceleration of the man. I inserted the equation with x=1.2, x0=0, v0=0.82, a=??, t=0.193 to get:

a= (x-x0- v0*t )/ (1/2*t^2)
a= (1.2 - 0.82*0.193) / (0.5 * 0.193^2)
a=57.71 m/s2 which is wrong according to the answer :(

2. Mar 14, 2017

### BvU

Hello Cynthia,

I think you have a little more time to catch up with the critter: you start 0.91 m behind him (her?) and he/she still has 1.20 m to go before escaping.

Help yourself with a sketch of the situation: time horizontal, position in the y direction

3. Mar 14, 2017

### CWatters

That's the problem. The roach starts 1.2m away from the counter. The man starts 1.2m + 0.91m away from the counter.

4. Mar 14, 2017

### Cynthia

Thank you so much BvU and CWatters I get it now

5. Mar 14, 2017

### kuruman

If you calculate the final speed of "you" just before you reach the counter, it is about 10 miles per hour! You are likely to be knocked unconscious, lying on the floor of a cheap motel with large roaches scurrying around.