We will use this equation:
s = 1/2*a*t^2 + v0*t + s0
where:
s = space traveled
a = acceleration
t = time
v0 = initial speed
s0 = initial space
In this case::
v0 = 0
s0 = 0
So our equation will look like that now:
s = 1/2 * a * t^2
let's calculate the acceleration first of all:
a = (vf - vi) / t
where vf is the final speed and vi is the initial speed. t is the time.
a = (25m/s) / 10s = 2.5 m/s^2
Now we can calculate the space:
s = 1/2 * (2.5 m/s^2) * (10s)^2 = 125m
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Hope it was helpful! Have a great day.
Answer:
Explanation:
Given
Initial speed 
distance traveled before coming to rest 
using equation of motion
where v=final velocity
u=initial velocity
a=acceleration
s=displacement
for 
using same relation we get
divide 1 and 2 we get
So a distance if 213.32 ft is required to stop the vehicle with 80 mph speed
C and c infeijnveirnvinefine
Answer:
It would take the object 5.4 s to reach the ground.
Explanation:
Hi there!
The equation of the height of a free-falling object at any given time, neglecting air resistance, is the following:
h = h0 + v0 · t + 1/2 · g · t²
Where:
h = height of the object at time t.
h0 = initial height.
v0 = initial velocity.
g = acceleration due to gravity (-32.2 ft/s² considering the upward direction as positive).
t = time
Let´s supose that the object is dropped and not thrown so that v0 = 0. Then:
h = h0 + 1/2 · g · t²
We have to find the time at which h = 0:
0 = 470 ft - 1/2 · 32.2 ft/s² · t²
Solving for t:
-470 ft = -16.1 ft/s² · t²
-470 ft / -16.1 ft/s² = t²
t = 5.4 s
Speed = wavelength * Frequency
s = 247 /s * 1.4 m
s = 345.8 m/s
In short, Your Answer would be 345.8 m/s
Hope this helps!
