Question

Big cockroaches can run as fast as 1.50 m/s over short distances. Suppose you turn on the light in an inexpensive motel and see one run directly away from you at a constant speed of 1.50 m/s. If you run after it to squish it beginning 0.80 m behind it with an initial velocity of 0.70 m/s, what minimum acceleration must you have to catch it before it reaches safety 1.30 m away from where it starts running?

Answer 1

a = 4.054 m/ s2

Step-by-step explanation:

`safety\ Time\ for\ cockroach = (Distance\ travelled\ by\ cockroach)/( speed of cockroach)`

`t = (1.3)/(1.5)`

t = 0.86 seconds

total distance travelled by you = 0.8 + 1.3 = 2.1 m

Initial velocity u = 0.7 m/s

let a is acceleration,

from equation of motion we have

s = ut + 0.5 at^2

`2.1 = 0.7 (1.3)/(1.5) + 0.5 a [(1.3)/(1.5)]^2`

2.1 = 0.60 + 0.37 a

1. 5 = 0.37a

a = 4.054 m/ s2