Answer:
C
Explanation:
i could be wrong but it seems the most logical
Explanation:
1.
We use the equation
h = 
, where
h is the height traveled,
g is the acceleration due to gravity and
t is the time taken to reach height h.
We can now calculate t to be
= 0.495 s
Let v be the initial velocity of the player.
The player deaccelarates from v m/s to 0 m/s in 0.495 s at the rate of 9.81 m/s^2.
v = 9.81 m/s^2 x 0.495 s = 4.85 m/s
2.
The player takes 0.3 s to increase his velocity from 0 m/s to 4.85 m/s. So his average accelaration is
4.85 m/s / 0.3 s = 16.2 m/s^2
Answer:
45.89m/s²
Explanation:
Given
Distance S = 305m
Time t = 3.64s
To get the acceleration during this run, we will apply the equation of motion:
S = ut+1/2at²
Substitute the given parameters into the formula and calculate the value of a
305 = 0+1/2 a(3.64)²
304 = 1/2(13.2496)a
304 = 6.6248a
a = 304/6.6248
a = 45.89m/s²
Hence the average acceleration during this run is 45.89m/s²
Answer:
1.125m/s^2
Explanation:
Since acceleration is defined as the rate of change in velocity with respect to time. Mathematically
v^2= u^2+2as
Where a,v,u and s are the acceleration, final velocity, initial velocity and distance respectively.
a = ?
u = 0m/s
v = 15m/s
s = 100m
Substituting the values into the formula above
v^2= u^2+2as
15^2=0^2+2×a×100
225= 0+200a
225= 200a
Divide both sides by 200
225/200 = 200a/200
a= 1.125m/s^2
Hence the acceleration of the car is 1.125m/s^2.
Note that the car accelerated uniformly from rest, that was why the initial velocity was 0m/s
