With that information you can only suppose a uniformly accelerated motion. This is, acceleration is constant.
Then, acceleration = change in velocity / change in time = (58 -54)km/h / 2 h = 4km/h / 2 h = 2 km/h^2
Then the equation for velocity, V is
V = Vo + a*t = Vo + 2 (km/h^2) * t = Vo + 2t
Vo is the initial velocity, which you can find using V = 54km/h and t = -2
Vo = V after 2 hours - a*(2hours) = 54km/h - 2(km/h^2)*2h = 54km/k - 4km/h = 50km/h
Then, the equation is: V = 50 km/h + 2t
Valid for constant acceleration.
It is 3.2 because you add 2.3 and 4.1 then divide your answer by 2.
<h2>
Option 3, 216 m is the correct answer.</h2>
Explanation:
We have initial velocity, u = 15 m/s
Time, t = 12 seconds
Final velocity, v = 21 m/s
We have equation of motion v = u + at
Substituting
21 = 15 + a x 12
a = 0.5 m/s²
Now we have equation of motion v² = u² + 2as
21² = 15² + 2 x 0.5 x s
s = 216 m
Displacement = 216 m
Option 3, 216 m is the correct answer.
Answer:
The time for final 15 cm of the jump equals 0.1423 seconds.
Explanation:
The initial velocity required by the basketball player to be able to jump 76 cm can be found using the third equation of kinematics as

where
'v' is the final velocity of the player
'u' is the initial velocity of the player
'a' is acceleration due to gravity
's' is the height the player jumps
Since the final velocity at the maximum height should be 0 thus applying the values in the above equation we get

Now the veocity of the palyer after he cover'sthe initial 61 cm of his journey can be similarly found as

Thus the time for the final 15 cm of the jump can be found by the first equation of kinematics as

where symbols have the usual meaning
Applying the given values we get
