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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,
Time,
Acceleration to top speed,
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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