A cheetah spotted a Gazelle. The cheetah runs at its top speed of 30 m/s for 15 seconds. Durning this time , a gazelle ,160 m from the runni

Question

A cheetah spotted a Gazelle. The cheetah runs at its top speed of 30 m/s for 15 seconds. Durning this time , a gazelle ,160 m from the running cheetah notices the danger ahead. The gazelle accelerated at 4.5m/s^2 for 6.0s, then continued running at a constant speed.

A) how far does the cheetah travel during 15s duration
B) how fast does the gazelle reach after 6.0 s

C) how far does the gazelle travel during the first 6.0 s and last 9.0
D) who’s ahead at the end of 15secs

in progress 0
Minh Khuê 2 months 2021-07-26T10:58:01+00:00 1 Answers 1 views 0

Answers ( )

    0
    2021-07-26T10:59:52+00:00

    Answer:

    A) 450 m

    B) 27 m/s

    C) 81 m, 243 m

    D) Gazelle

    Explanation:

    A)

    Since, the Cheetah is running at constant speed. Therefore, we use the equation:

    s₁ = v₁t

    where,

    s₁ = distance covered by Cheetah = ?

    v₁ = speed of Cheetah = 30 m/s

    t = time taken = 15 s

    Therefore,

    s₁ = (30 m/s)(15 s)

    s₁ = 450 m

    B)

    For final speed of Gazelle at the end of 6 s acceleration time we use 1st equation of motion:

    Vf = Vi + at

    where,

    Vf = Final Speed of Gazelle at the end of 6 s = ?

    Vi = Initial Speed of Gazelle = 0 m/s (Since, Gazelle is initially at rest)

    t = time taken = 6 s

    a = acceleration = 4.5 m/s²

    Therefore,

    Vf = (0 m/s) + (4.5 m/s²)(6 s)

    Vf = 27 m/s

    C)

    For the distance covered by Gazelle at the end of 6 s acceleration time we use 2nd equation of motion:

    s₂ = Vi t + (0.5)at²

    where,

    Vi = Initial Speed of Gazelle = 0 m/s (Since, Gazelle is initially at rest)

    t = time taken = 6 s

    a = acceleration = 4.5 m/s²

    Therefore,

    s₂ = (0 m/s)(6 s) + (0.5)(4.5 m/s²)(6 s)²

    s₂ = 81 m

    Now, for the distance covered during the last 9 s at constant velocity Vf, we use equation:

    s₃ = Vf t

    where,

    s₃ = distance covered by Gazelle in last 9 s = ?

    t = time = 9 s

    Therefore:

    s₃ = (27 m/s)(9 s)

    s₃ = 243 m

    D)

    We know that, at the end of 15 seconds:

    Distance covered by Cheetah = s₁ = 450 m

    Distance Covered by Gazelle = s₂ + s₃ = 81 m + 243 m = 324 m

    If we take the initial position of Cheetah as origin. Then the positions of both Gazelle and Cheetah with respect to origin will be:

    Position of Cheetah = 450 m ahead of origin

    Position of Gazelle = 324 m + 160 m = 484 m (Since, Gazelle was initially 160 m ahead of Cheetah)

    Hence, it is clear that Gazelle is ahead at the end of 15 s.

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )