When riding a bicycle, if you stop pedaling you will still continue to move forward due to inerita.
The solution would be like
this for this specific problem:
<span>
The force on m is:</span>
<span>
GMm / x^2 + Gm(2m) / L^2 = 2[Gm (2m) / L^2] ->
1
The force on 2m is:</span>
<span>
GM(2m) / (L - x)^2 + Gm(2m) / L^2 = 2[Gm (2m) / L^2]
-> 2
From (1), you’ll get M = 2mx^2 / L^2 and from
(2) you get M = m(L - x)^2 / L^2
Since the Ms are the same, then
2mx^2 / L^2 = m(L - x)^2 / L^2
2x^2 = (L - x)^2
xsqrt2 = L - x
x(1 + sqrt2) = L
x = L / (sqrt2 + 1) From here, we rationalize.
x = L(sqrt2 - 1) / (sqrt2 + 1)(sqrt2 - 1)
x = L(sqrt2 - 1) / (2 - 1)
x = L(sqrt2 - 1) </span>
= 0.414L
<span>Therefore, the third particle should be located the 0.414L x
axis so that the magnitude of the gravitational force on both particle 1 and
particle 2 doubles.</span>
Answer: The Porsche Wins. Arrives 19secs earlier before the Honda.
Explanation: The head start of 1secs corresponded with the difference in acceleration {3.5m/s² - 3m/s²}=0.5m/s²
Using the first equation of motion we obtain the velocity which correspond to this acceleration (0.5m/s²) and time of 1secs where initial velocity u = 0
V = u + at
V= 0 + at
V = 0.5 * 1 = 0.5m/s
Now let find the velocity of each of the car.
If V. a
0.5m/s 0.5m/s²
........m/s. 3.5m/s²
Velocity of porche Vp
= (3.5/0.5) * 0.5 = 3.5 m/s
Also if. V. a
0.5m/s. 0.5m/s²
.....m/s. 3 m/s²
Velocity of Honda Vh
= {3/0.5} * 0.5 = 3m/s
So let's find the time t taken by both cars to cover the distance of 400m
Recall,
Velocity = distance/time
Time t = distance/velocity
For Porche, t = 400/3.5 =114.29secs
For Honda, t = 400/3 = 133.33secs
Looking critically, we noticed that Porche car took shorter time.
The difference In time is
= (133.33 - 114.29)secs = 19.04secs
