Answer:
<h2>32m/s^2</h2>
Explanation:
We want to find the acceleration based on the given data
Given
distance s=400m
time t= 5s
u= 0m/s since it started from rest
We apply the following expression
s=ut+1/2at^2
substituting we have
400=0*5+1/2*a(5)^2
400=25a/2
cross multiply
25a=400*2
25a=800
divide both sides by 25 we have
a=800/25
a=32m/s^2
The acceleration is 32m/s^2
I believe the answer is B)
Hope this helps*
Maintaining an orbit has nothing to do with the satellite's speed.
ANY (tangential) speed is enough to stay in orbit, if the satellite
just stays away from Earth's atmosphere.
If the satellite completely stops moving 'sideways' at all, and just
hangs there, then the forces of gravity between the satellite and
the Earth will pull them together ... the satellite will fall into the
atmosphere and then to the ground.
Answer:
<h2>a)
Acceleration is 3.09 m/s²</h2><h2>
b) Distance traveled is 45.05 m</h2><h2>c)
Time taken to travel 250 m is 12.72 s</h2>
Explanation:
a) We have equation of motion v = u + at
Initial velocity, u = 0 km/hr = 0 m/s
Final velocity, v = 60 km/hr = 16.67 m/s
Time, t = 5.4 s
Substituting
v = u + at
16.67 = 0 + a x 5.4
a = 3.09 m/s²
Acceleration is 3.09 m/s²
b) We have equation of motion s = ut + 0.5 at²
Initial velocity, u = 0 m/s
Acceleration, a = 3.09 m/s²
Time, t = 5.4 s
Substituting
s = ut + 0.5 at²
s = 0 x 5.4 + 0.5 x 3.09 x 5.4²
s = 45.05 m
Distance traveled is 45.05 m
c) We have equation of motion s = ut + 0.5 at²
Initial velocity, u = 0 m/s
Acceleration, a = 3.09 m/s²
Displacement, s = 0.25 km = 250 m
Substituting
s = ut + 0.5 at²
250 = 0 x t + 0.5 x 3.09 xt²
t = 12.72 s
Time taken to travel 250 m is 12.72 s
