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Marrrta [24]
3 years ago
7

Suppose a cheetah running at a velocity of 5 m/s east slows down. After 15 seconds, the cheetah has reached a velocity of 25 m/s

. What is the
cheetah's acceleration?
Physics
1 answer:
Darina [25.2K]3 years ago
5 0

Answer:

1.33m/s²

Explanation:

Given parameters:

Initial velocity  = 5m/s

Final velocity  = 25m/s

Time taken  = 15seconds

Unknown:

Acceleration  = ?

Solution:

Acceleration is the rate of change of velocity with time.

  Acceleration  = \frac{v - u}{t}  

  v is the final velocity

  u is the initial velocity

  t is the time taken

 So;

    Acceleration  = \frac{25 - 5}{15}    = 1.33m/s²

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Determine the velocity that a car should have while traveling around a frictionless curve of radius 100m and that is banked 20 d
alex41 [277]

Answer:

v=18.89\frac{m}{s}

Explanation:

From the free-body diagram for the car, we have that the normal force has a vertical component and a horizontal component, and this component act as the centripetal force on the car:

\sum F_x:Nsin(20^\circ)=ma_c(1)\\\sum F_y:Ncos(20^\circ)=mg(2)

Solving N from (2) and replacing in (1):

N=\frac{mg}{cos(20^\circ)}\\(\frac{mg}{cos(20^\circ)})sin(20^\circ)=ma_c\\gtan(20^\circ)=a_c

The centripetal acceleration is given by:

a_c=\frac{v^2}{r}

Replacing and solving for v:

gtan(20^\circ)=\frac{v^2}{r}\\v=\sqrt{grtan(20^\circ)}\\v=\sqrt{9.8\frac{m}{s^2}(100m)tan(20^\circ)}\\v=18.89\frac{m}{s}

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