A lunar lander is descending toward the moon's surface. Until the lander reaches the surface, its height above the surface of the moon is given by y\left( t \right) = b - ct + dt^2, where b = 740 m is the initial height of the lander above the surface, c = 64.0 m/s, and d = 1.03 m/s^2.What is the velocity of the lander just before it reaches the lunar surface? please answer quikly.
A lunar lander is descending toward the moon's surface. Until the lander reaches the surface, its height above the surface of the moon is given by y(t) = b - ct + dt^2, where b = 750 m is the initial height of the lander above the surface, c = 65.0 m/s, and d = 1.03 m/s^2.
What is the velocity of the lander just before it reaches the lunar surface?
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Best Answer
Dr Zorro answered 3 years ago
The time when the lunar lander reaches the surface is found by solving y(t)=0.
b - ct + d t^2 = 0 => t = (c +- sqrt( c^2 - 4 b d)) / (2d)
t = (65.0 m/s +- sqrt((65.0m/s)^2 - 4 * 750 m * 1.03 m/s^2))/(2*1.03 m/s^2)
t = (65.0 m/s +- 33.7 m/s) / (2.6 m/s^2)
t = 12.0 s or t = 38.0 s .
Obviously we need the smallest time solution (when y equals zero from the first time, the second solution being non-physical as it is the solution corresponding to the lunar lander going down through the surface and coming back up to y=0 from under the surface)
So y=0 when t=12.0 s
The velocity can now be found from:
v(t) = dy/dt = -c + 2 d t
So at the surface we have
v(12.0s) = -65.0 m/s + 2 * 1.03 m/s^2 * 12.0s
= -40.3 m/s
I.e. a downward speed of 40.3 m/s (Although correctly calculated, this seems a bit large to me as it should be a lunar lander, not a lunar crasher..!)
Well, well, well, descending towards the moon's surface, huh? Talk about a risky business! Now, let's get to the question at hand. To find the velocity of the lander just before it reaches the surface, we need to take the derivative of the height function with respect to time.
So, the velocity function, v(t), is obtained by differentiating y(t) = b - ct + dt^2 with respect to t. Let's give it a go:
v(t) = -c + 2dt
Now, since we want to find the velocity just before the lander reaches the surface, we need to substitute t with the value that makes y(t) equal to zero. In other words, we solve y(t) = 0 for t.
0 = b - ct + dt^2
Since we're looking for the velocity just before reaching the lunar surface, we'll consider the positive root of the equation. After solving for t, we substitute this value of t into the velocity function, v(t), to get the velocity just before landing.
And there you go, my friend! You have successfully obtained the velocity of the lunar lander just before it kisses the moon's surface! Now just remember to buckle up for a soft landing. Good luck!
To find the velocity of the lander just before it reaches the lunar surface, we need to take the derivative of the equation y(t) with respect to time.
For the given equation: y(t) = b - ct + dt^2
Differentiating y(t) with respect to t:
dy/dt = -c + 2dt
Substitute the given values:
dy/dt = -64 + 2(1.03)t
To find the velocity just before it reaches the lunar surface, we need to find the value of t when the height y(t) is zero.
0 = b - ct + dt^2
0 = 740 - 64t + 1.03t^2
Solving this quadratic equation, we find t = 29.7 seconds (rounded to one decimal place).
Now, substitute the value of t into the velocity equation:
dy/dt = -64 + 2(1.03)(29.7)
Calculating this, the velocity of the lander just before it reaches the lunar surface is approximately -108.78 m/s (rounded to two decimal places). Note that the negative sign indicates a downward velocity.
To find the velocity of the lander just before it reaches the lunar surface, we need to find the derivative of the height function with respect to time.
Given:
y(t) = b - ct + dt^2
To find the derivative, we differentiate each term separately:
dy/dt = d(b - ct + dt^2)/dt
The derivative of a constant (b) with respect to time is zero, since it does not change with time.
dy/dt = -c + 2dt
Now we have the velocity of the lander at any given time, which is the derivative of the height function. However, we are interested in finding the velocity just before the lander reaches the lunar surface, which means t = 0.
To find the velocity just before the lander reaches the surface, substitute t = 0 into the velocity function:
v = dy/dt
v = -c + 2d(0)
v = -c
So, the velocity of the lander just before it reaches the lunar surface is -c, which in this case is -64.0 m/s.