A cheetah starting form rest can reach a maximum speed of 28m.s in 2s and it can maintain this maxium speed for 15s after this it must rest.

suppose an antelope and a cheetah initially at rest are separated by a displcament of magnitude 100m.The antelope and cheetah see each other simultaneously and then start running westwards immediately.

1.1Calculate the acceleration of the cheetah in the first 2s.

1.2 Calculate the maximum distance that the cheetah can run before it comes to rest.

1.3 If the acceleration of the antelope for the first 2s of the chase in 12.5 m.s2 , determine whether the cheetah will catch the antelope

To find the answers to these questions, we need to consider the motion equations for constant acceleration.

1.1 To calculate the acceleration of the cheetah in the first 2 seconds, we can use the equation:

acceleration = (final velocity - initial velocity) / time

Here, the initial velocity is 0 m/s (since it starts from rest), the final velocity is 28 m/s, and the time is 2 seconds. Plugging in these values, we get:

acceleration = (28 m/s - 0 m/s) / 2 s
acceleration = 28 m/s / 2 s
acceleration = 14 m/s²

Therefore, the acceleration of the cheetah in the first 2 seconds is 14 m/s².

1.2 To calculate the maximum distance that the cheetah can run before it comes to rest, we need to find the distance traveled during the time it maintains its maximum speed.

The distance traveled at a constant speed can be calculated using the equation:

distance = initial velocity * time

Here, the initial velocity is 28 m/s (the maximum speed reached by the cheetah), and the time is 15 seconds. Plugging in these values, we get:

distance = 28 m/s * 15 s
distance = 420 m

Therefore, the maximum distance that the cheetah can run before it comes to rest is 420 meters.

1.3 To determine whether the cheetah will catch the antelope, we need to compare the distances each can cover during their respective accelerations.

The distance traveled during acceleration can be calculated using the equation:

distance = (initial velocity * time) + (0.5 * acceleration * (time^2))

For the cheetah, the initial velocity is 0 m/s, the time is 2 seconds, and the acceleration is 14 m/s². Plugging in these values, we get:

distance_cheetah = (0 m/s * 2 s) + (0.5 * 14 m/s² * (2 s)²)
distance_cheetah = 0 m + (0.5 * 14 m/s² * 4 s²)
distance_cheetah = 0 m + 28 m²/s²
distance_cheetah = 28 m

For the antelope, the initial velocity is 0 m/s, the time is 2 seconds, and the acceleration is 12.5 m/s². Plugging in these values, we get:

distance_antelope = (0 m/s * 2 s) + (0.5 * 12.5 m/s² * (2 s)²)
distance_antelope = 0 m + (0.5 * 12.5 m/s² * 4 s²)
distance_antelope = 0 m + 25 m²/s²
distance_antelope = 25 m

Since the distance traveled by the cheetah (28 m) is greater than the distance traveled by the antelope (25 m) during the first 2 seconds of the chase, the cheetah will catch the antelope.