How long for a cop car accelerating at 9 km/h² to catch a speeder going 110 km/h?

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anglum said:
so then for the horizontal distance it is just 17 cos 50 (2.62)?

yes.
 
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ok 2 more problems if u don't mind?

#1

a rotating cylinder 10 miles long and 4.9 miles in radius is in space...

acceleration of gravity is 9.8m/s squared

what angular speed must the cylinder have so that the centripetal acceleration at its surface equals the free fall acceleration on earth?
 
ok i got that problem on my own actually WOOO HOOO

now the 2nd problem if u don't mind
 
014 (part 1 of 2) 10 points
The speed of a point on a rotating turntable,
which is 0.131 m from the center, changes at a
constant rate from rest to 0.958 m/s in 1.95 s.
At t1 = 1.6 s, find the magnitude of the
tangential acceleration. Answer in units of
m/s2.
015 (part 2 of 2) 10 points
At t1 = 1.6 s, ¯find the magnitude of the total
acceleration of the point. Answer in units of
m/s2.
 
Last edited:
learning these are my last 2 problems ... sorry to keep bothering you
 
anglum said:
014 (part 1 of 2) 10 points
The speed of a point on a rotating turntable,
which is 0.131 m from the center, changes at a
constant rate from rest to 0.958 m/s in 1.95 s.
At t1 = 1.6 s, find the magnitude of the
tangential acceleration. Answer in units of
m/s2.

This is just the change in velocity divided by time. ie 0.958/1.95.

015 (part 2 of 2) 10 points
At t1 = 1.6 s, ¯find the magnitude of the total
acceleration of the point. Answer in units of
m/s2.

What is the centripetal acceleration?
 
anglum said:
ok 2 more problems if u don't mind?

#1

a rotating cylinder 10 miles long and 4.9 miles in radius is in space...

acceleration of gravity is 9.8m/s squared

what angular speed must the cylinder have so that the centripetal acceleration at its surface equals the free fall acceleration on earth?

use centripetal acceleration = v^2/r. At a radius r, v = rw.