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Rudik [331]
2 years ago
6

If a rock is thrown upward on the planet mars with a velocity of 11 m/s, its height (in meters) after t seconds is given by h =

11t − 1.86t2. (a) find the velocity of the rock after two seconds. 3.56 m/s (b) find the velocity of the rock when t =
a. 11−3.72a m/s (c) when will the rock hit the surface? (round your answer to one decimal place.) t = s (d) with what velocity will the rock hit the surface? m/s
Physics
1 answer:
Butoxors [25]2 years ago
3 0
(a) 3.56 m/s 
(b) 11 - 3.72a 
(c) t = 5.9 s 
(d) -11 m/s  
For most of these problems, you're being asked the velocity of the rock as a function of t, while you've been given the position as a function of t. So first calculate the first derivative of the position function using the power rule. 
y = 11t - 1.86t^2 
y' = 11 - 3.72t 
Now that you have the first derivative, it will give you the velocity as a function of t. 
(a) Velocity after 2 seconds. 
y' = 11 - 3.72t 
y' = 11 - 3.72*2 = 11 - 7.44 = 3.56 
So the velocity is 3.56 m/s  
(b) Velocity after a seconds. 
y' = 11 - 3.72t 
y' = 11 - 3.72a  
So the answer is 11 - 3.72a  
(c) Use the quadratic formula to find the zeros for the position function y = 11t-1.86t^2. Roots are t = 0 and t = 5.913978495. The t = 0 is for the moment the rock was thrown, so the answer is t = 5.9 seconds.  
(d) Plug in the value of t calculated for (c) into the velocity function, so: 
y' = 11 - 3.72a
 y' = 11 - 3.72*5.913978495
 y' = 11 - 22
 y' = -11 
 So the velocity is -11 m/s which makes sense since the total energy of the rock will remain constant, so it's coming down at the same speed as it was going up.
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a car traveling at 28 m/s slows down at a constant rate for 4 seconds until it stops. what is its acceleration?
astra-53 [7]
We are given the following conditions:
v_{0} = 28m/s

t = 4s

v_{f} = 0m/s

Since we are told it slows down at a constant rate, we know we are dealing with uniformed accelerated motion, or UAM for short.

So we look at our kinematics equation that contains the given variables.

v_f = v_0 + at

This contains our target variable and our initial conditions. Plug and chug, we get:

(0m/s) = (28m/s) + a(4s) -28m/s = (4s)*a a = -7m/s^2

Thus the answer is:

a = -7m/s^2

Hope this helps!


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Paraphin [41]

Answer:

I believe it's 8.09 seconds, but I'm rusty on my physics.

Explanation:

The equation for solving the time it takes for an object to fall is \sqrt{2d/g}

So multiply the distance times 2, and you get 642 meters. Then you divide by gravities acceleration constant, 9.8, and you get 65.51. Finally, \sqrt{65.51}, and you get 8.09 seconds.

I pulled the equation off of wikipedia and I'm unsure if it's the correct one, so hopefully this is correct. :/

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