Step-by-step explanation:
we know from the laws of motion that in the equation
h = 20t - 1.86t²
the gravitational acceleration is in the factor of the t² term.
h = v0t + 1/2 × g × t²
v0 being the initial velocity (20 m/s).
and we can therefore see, that the gravitational acceleration "a" on Mars is 2×1.86 = 3.72 m/s²
(a)
the velocity v after 2 seconds is (first law of motion)
v = v0 + at = 20 - 3.72×2 = 12.56 m/s
gravity pulling down, so negative acceleration.
(b)
first we need the time when the rock is at 25 m.
25 = 20t - 1.86t²
0 = -1.86t² + 20t - 25
the solution to a quadratic equation is always
x = (-b ± sqrt(b² - 4ac))/(2a)
in our case
x = t
a = -1.86
b = 20
c = -25
t = (-20 ± sqrt(400 - 4×-1.86×-25))/(2×-1.86) =
= (-20 ± sqrt(214))/-3.72
t1 = (-20 + 14.62873884...)/-3.72 =
= 1.443887409... s ≈ 1.44 s
t2 = (-20 - 14.62873884...)/-3.72 =
= 9.308800763... s ≈ 9.31 s
that means, on its way up the rock reached 25m after
1.44 s.
on its way down the rock reached 25 m after
9.31 s.
velocity 0 (the rock came to a stop before falling back down) was after
0 = 20 - 3.72t
3.72t = 20
t = 20/3.72 = 5.376344086... s
that means the rock was falling after this point back to the ground and was gaining speed again.
that accelerating phase until the rock was again at 25 m was
9.308800763... - 5.376344086... = 3.932456677... s
long
the velocity at these points was
v-up = 20 - 3.72×1.443887409... = 14.62873884... m/s ≈
≈ 14.63 m/s
v-down = 0 + 3.72×3.932456677... = 14.62873884... m/s ≈
≈ 14.63 m/s
as expected : the rock did a perfectly symmetric flight path between the two 25 m marks.
so time and speed on both sides had to be identical.
but we have proven it.
Answer:
a) 12.56 m/s
b) up: 14.63 m/s (2 d.p.)
down: -14.63 m/s (2 d.p.)
Step-by-step explanation:
[tex]\boxed{\begin{minipage}{9 cm}\underline{The Constant Acceleration Equations (SUVAT)}\\\\s = displacement in m (meters)\\u = initial velocity in m s$^{-1}$ (meters per second)\\v = final velocity in m s$^{-1}$ (meters per second)\\a = acceleration in m s$^{-2}$ (meters per second per second)\\t = time in s (seconds)\\\\When using SUVAT, assume the object is modeled\\ as a particle and that acceleration is constant.\end{minipage}}[/tex]
The average gravitational acceleration on Mars is 3.721 m/s² (about 38% of that of Earth). The fact that the rock is thrown from the surface of Mars rather than the surface of Earth is of no consequence when using the equations of constant acceleration (SUVAT equations) as long as acceleration is not used in the calculations.
Given:
[tex]s=20t-1.86t^2, \quad u=20, \quad t=2[/tex]
[tex]\begin{aligned}\textsf{Using }\: s&=\dfrac{1}{2}(u+v)t\\\\\implies 20(2)-1.86(2)^2 & = \dfrac{1}{2}(20+v)(2)\\\\32.56 & = 20+v\\\\v & = 32.56-20\\\\v & = 12.56\:\: \sf m/s\end{aligned}[/tex]
Therefore, the velocity of the rock after 2 s is 12.56 m/s.
Find the time when the height of the rock is 25 m:
[tex]\begin{aligned}20t-1.86t^2 & = s\\\implies 20t-1.86t^2 & = 25\\1.86t^2-20t+25 & = 0\end{aligned}[/tex]
Quadratic Formula
[tex]x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0[/tex]
Therefore:
[tex]a=1.86, \quad b=-20, \quad c=25[/tex]
[tex]\implies t=\dfrac{-(-20) \pm \sqrt{(-20)^2-4(1.86)(25)} }{2(1.86)}[/tex]
[tex]\implies t=\dfrac{20 \pm \sqrt{214}}{3.72}[/tex]
[tex]\implies t=9.308800763..., 1.443887409...[/tex]
Therefore, the height of the rock is 25 m when:
To find the velocity (v) of the rock at these times, substitute the found values of t into the equation, along with s = 25 and u 20 m/s:
[tex]\begin{aligned}\textsf{Using }\: s&=\dfrac{1}{2}(u+v)t\\\\\implies v & = \dfrac{2s}{t}-u\\\\v & = \dfrac{2(25)}{t}-20\\\\v & = \dfrac{50}{t}-20\\\\\end{aligned}[/tex]
When t = 1.443887409...s
[tex]\implies v=\dfrac{50}{1.443887409...}-20=14.63\:\: \sf m/s \: \:(2\:d.p.)[/tex]
When t = 9.308800763...s
[tex]\implies v=\dfrac{50}{9.308800763...}-20=-14.63\:\: \sf m/s \: \:(2\:d.p.)[/tex]
Therefore, the velocity of the rock when its height is 25 m is:
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