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?

top= 66 mi/h, accel = 3.71 s, d = 54.8 m, t total = 6.56 s

Explanation:

Here is the complete question.

Cheetahs can accelerate to a speed of 19.5 m/s in 2.45 s and can continue to accelerate to reach a top speed of 29.5 m/s . Assume the acceleration is constant until the top speed is reached and is zero thereafter. Let the + direction point in the direction the cheetah runs. Express the cheetah's top speed top in miles per hour (mi/h) . top= mi/h Starting from a crouched position, how much time accel does it take a cheetah to reach its top speed and what distance does it travel in that time? accel= s = m If a cheetah sees a rabbit 139.0 m away, how much time total will it take to reach the rabbit, assuming the rabbit does not move and the cheetah starts from rest? total= s

1. The cheetah's top speed in miles per hour = 29.5/1609 × 3600  = 66 mi/h

2. From the crouching position to top speed, the cheetah first accelerates to a speed v₁ = 19.5 m/s from rest in 2.45 s. To calculate this acceleration, we use Newton's equation of motion at constant acceleration v₁ = v₀ + at₀

v₀ = 0, v₁ = 19.5 m/s², t₀ = 2.45 s.

a = v₁ - v₀/t₀ =  (19.5 - 0)/2.45 = 7.959 m/s ≅ 7.96 m/s².

It then continues at this acceleration until it reaches a top speed of 29.5 m/s. The time required for this using v₂ = v₁ + at₁ ⇒ t₁ = v₂ - v₁/a

v₁ = 19.5 m/s, v₂ = 29.5 m/s, a = 7.96 m/s².

t₁ = (v₂ - v₁)/a = (29.5 -19.5)/7.96 = 10/7.96 = 1.256 s ≅ 1.26 s. So. from a crouching position to top speed, the time equals t = t₀ + t₁ = 2.45 + 1.26 = 3.71 s. So t accel = 3.71 s

The distance, d it travels in this time of 3.71 s is using d = ut + 1/2at²

u = 0, t = 3.71 s, a = 7.96 m/s²

d = 0 × 3.71 + 1/2 × 7.96 × 3.71² = 0 + 54.78 = 54.78 m ≅ 54.8 m

From rest to top speed, the cheetah takes 3.71 s at a distance of 54.8 m. The rabbit is 139.0 m away. So, the distance left to be covered is 139.0 - 54.8 m = 84.2 m. This distance is covered at top speed since the cheetah stops accelerating when it gets to top speed. So, the time t₂ covered in this distance is t₂ = 84.2/29.5 = 2.85 s. So, the total time t total = time to reach to speed from rest + time to reach rabbit from top speed = 3.71 + 2.85 = 6.56 s

magnityde and direction