A car is traveling due north at 23.6 m>s.
Find the velocity of the car after 7.10 s if its
acceleration is
The acceleration is known to be: a(t) = 1.7 m/s2.
We must integrate over time to obtain the velocity function, and the results are:
v(t) = (1.7m/s^2)
*t + v0
If we suppose that we begin at 23.6 m/s, then the initial velocity is: v0 = 23.6 m/s, where v0 is the beginning velocity.
The velocity formula is then: v(t) = (1.7m/s2).
*t + 23.6 m/s
We now seek to determine the value of t such that v(t) = 27.8 m/s.
Consequently, v(t) = 27.8 m/s = (1.7 m/s2)
*t + 23.6 m/s = (1.7 m/s2) 27.8 m/s - 23.6 m/s
t = 2.5 seconds when *t 4.2 m/s = (1.7 m/s/2)
At such acceleration, 2.5 seconds are required.
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You can determine wavelength with “x”
It's a projectile near the earth under the influence of gravity only.
It might make more sense putting it another way but this is basically it. you just take the minutes and divide them by 60 to convert them to hours. then simplify the ratio
I have a strange hunch that there's some more material or previous work
that goes along with this question, which you haven't included here.
I can't easily find the dates of Mercury's extremes, but here's some of the
other data you're looking for:
Distance at Aphelion (point in it's orbit that's farthest from the sun):
<span><span><span><span><span>69,816,900 km
0. 466 697 AU</span>
</span>
</span>
</span>
<span>
Distance at Perihelion
(</span></span><span>point in it's orbit that's closest to the sun):</span>
<span><span><span><span>46,001,200 km
0.307 499 AU</span> </span>
Perihelion and aphelion are always directly opposite each other in
the orbit, so the time between them is 1/2 of the orbital period.
</span><span>Mercury's Orbital period = <span><span>87.9691 Earth days</span></span></span></span>
1/2 (50%) of that is 43.9845 Earth days
The average of the aphelion and perihelion distances is
1/2 ( 69,816,900 + 46,001,200 ) = 57,909,050 km
or
1/2 ( 0.466697 + 0.307499) = 0.387 098 AU
This also happens to be 1/2 of the major axis of the elliptical orbit.
