1) an antelope runs with a constant velocity of 22.0ms^-1 by a cheetah that is l

Question

1) an antelope runs with a constant velocity of 22.0ms^-1 by a cheetah that is lying in the grass. The cheetah gets up and begins chasing directly after the antelope just as it has gone past , with a constant acceleration of 8.60 ms^-1, until it reaches its stop speed od 30.0 ms^-1. At that point the cheetah runs with constant velocity. How far does the cheetah run before it catches up to the antelope?

2) a pendulum bob has an instantaneous speed of 3.00 ms^-1 when it is in the configuration shown in the diagram. Take r= 1.00 m and (tetha)= 15 degrees

a) what is the magnitude of the radial component of acceleration of the bob?
b)what is the magnitude of the tangential component acceleration of the bob?

c) what is the total acceleration (relative to the radial direction)?

Explanation / Answer

time taken to reach stop speed = 30/8.6 = 3.48 sec

distance travelled by antelope in this time = 76.744 m

now the relative speed between cheetah and antelope = 8 m/s

time taken for cheetah to travel this distance = 76.744/8 = 9.59 sec

total distance run by cheetah = 30 +76.44 = 106.44 mtrs

B) acceleration a = v^2/r = 9 m/sec

Magnitude of radial component of acceleration = acos(?) = 9*cos(15*) = 8.69

Magnitude of tangential component = asin(?) = 9*sin(15) = 2.329

Total acceleration ?ax^2+ay^2 = 9

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.