5-A cheetah can accelerate from rest to 26.0 m/s in 6.26 s. (a) Assuming constan

Question

5-A cheetah can accelerate from rest to 26.0 m/s in 6.26 s.
(a) Assuming constant acceleration, how far has the cheetah run in this
time?
(b) After sprinting for just 3.13 s, is the cheetah's speed 13.0 m/s, more
than 13.0 m/s, or less than 13.0 m/s?
(c) What is the cheetah's average speed for the first 3.13 s of its sprint?

For the second 3.13 s of its sprint?
(d) Calculate the distance covered by the cheetah in the first 3.13 s and the
second 3.13 s.
first 3.13 s
second 3.13 s

9-a car in free fall. How long would it take for the car to go from 0 to
10 mi/h?

show your work

Explanation / Answer

Given that,

vi = 0 and v1 = 26.0 m/s t = 6.26 s

(a)we know that by defination, acceleration is

a = v/t = 26.0/6.26 = 4.15 m/s2

from equation of motion:

D = vi t + 1/2 a t2

D = 0 + 0.5 x 4.15 x (6.26)2 = 81.3 meters

Hence, D = 81.3 meters

(b)In this case, t = 3.13 s

We know from first eqn of motion that,

v = vi + at = 0 + 4.15 x 3.13 = 13 m/s

Hence, cheetah's speed = 13.0 m/s(its equal to)

(c) V = 13.0 m/s

(d)distance covered at t = 3.13s

D = vi t + 1/2 a t2 =0+ 0.5 x 4.15 x (3.13)2= 0 + 20.33 = 20.33 m

distance covered for the second 3.13 s of the sprint will be(in this case, intial speed = 13m/s)

D' = vi t +  1/2 a t2 = 13 x 3.13 + 0.5 x 4.15 x (3.13)2= 40.7 + 20.33 = 61.03 m

Hence, the distance covered in the first sprint is 20.33 m and in second sprint is 61.03m.

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