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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?

Respuesta :

Answer:

a = 4.054 m/ s2

Explanation:

[tex]safety\ Time\ for\ cockroach = \frac{Distance\ travelled\ by\ cockroach}{ speed of cockroach}[/tex]

[tex]t = \frac{1.3}{1.5}[/tex]

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

[tex]2.1 = 0.7 \frac{1.3}{1.5} + 0.5 a [\frac{1.3}{1.5}]^2[/tex]

2.1 = 0.60 + 0.37 a

1. 5 = 0.37a

a = 4.054 m/ s2

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