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Cheetahs can accelerate to a speed of 20.0 m/s in 2.50 s and can continue to accelerate to reach a top speed of 30.0 m/s. Assume the acceleration is constant until the top speed is reached and is zero thereafter. Let the +x-direction point in the direction the cheetah runs. Express the cheetah’s top speed (vtop) in miles per hour (mi/h).
Starting from a crouched position, how much time (t-accel) does it take a cheetah to reach its top speed, and what distance (d) does it travel in that time?

Sagot :

Answer:

The cheetah’s initial velocity is 0 m/s. Let’s use the following equation to determine the acceleration.

Explanation:

Acceleration = (v/t) = (20/2.5) = 8m/sec^2.

1) Time to 29m/sec. = (v/a) = (29/8) = 3.625 secs. (3.6)

Distance covered to top speed = (v^2/2a) = 52.5625 metres (52.6).

2) Distance to run at top speed = (120 - 52.5625) = 67.4375 metres.

Time = (d/v) = (67.4375/29) = 2.325 secs.

Total time to warthog = (2.325 + 3.625) = 5.95 secs.

Cheetah's top speed will be "67.12 mi/h". A complete solution is provided below.

Given that:

Speed,

  • 20.0 m/s

Time,

  • 2.50 s

Acceleration to top speed,

  • 30.0 m/s

Now,

Acceleration (a) of cheetah will be:

= [tex]\frac{20}{2.5}[/tex]

= [tex]8 \ m/s^2[/tex]

Top speed ([tex]V_{top}[/tex]) of cheetah

= [tex]\frac{30}{1609}\times 3600[/tex]

= [tex]67.12 \ mi/h[/tex]

→ The time required (t) to reach the top speed will be:

= [tex]\frac{30}{8}[/tex]

= [tex]3.75 \ sec[/tex]

and,

→ The distance travelled (d) by cheetah will be:

= [tex]\frac{1}{2}\times 8\times (3.75)^2[/tex]

= [tex]\frac{1}{2}\times 8\times 14.0625[/tex]

= [tex]56.25[/tex]

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