What constant acceleration do you need to catch up?

[Cynthia]

Homework Statement

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 traveled 1.20 m, just short of safety under a counter?

Homework Equations

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

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 :(

Please help, use simple physics as I am very new to the subject. And thank you in advance !

 

[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

 

[CWatters]

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.

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

 

[Cynthia]

Thank you so much BvU and CWatters I get it now

 

[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.