(1)
Cheetah speed:
Its position at time t is given by
(1)
Gazelle speed:
the gazelle starts S0=96.8 m ahead, therefore its position at time t is given by
(2)
The cheetah reaches the gazelle when . Therefore, equalizing (1) and (2) and solving for t, we find the time the cheetah needs to catch the gazelle:
(2) To solve the problem, we have to calculate the distance that the two animals can cover in t=7.5 s.
Cheetah:
Gazelle:
So, the gazelle should be ahead of the cheetah of at least
Answer:
The skater 1 and skater 2 have a final speed of 2.02m/s and 2.63m/s respectively.
Explanation:
To solve the problem it is necessary to go back to the theory of conservation of momentum, specifically in relation to the collision of bodies. In this case both have different addresses, consideration that will be understood later.
By definition it is known that the conservation of the moment is given by:
Our values are given by,
As the skater 1 run in x direction, there is not component in Y direction. Then,
Skate 1:
Skate 2:
Then, if we applying the formula in X direction:
m_1v_{x1}+m_2v_{x2}=(m_1+m_2)v_{fx}
75*5.45-75*1.41=(75+75)v_{fx}
Re-arrange and solving for v_{fx}
v_{fx}=\frac{4.04}{2}
v_{fx}=2.02m/s
Now applying the formula in Y direction:
Therefore the skater 1 and skater 2 have a final speed of 2.02m/s and 2.63m/s respectively.