Motion with constant acceleration

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Homework Help Overview

The discussion revolves around two physics problems involving motion with constant acceleration. The first problem concerns a rocket's ascent and the calculation of its maximum height after its motor stops, while the second problem involves a motorist's reaction time to avoid hitting a deer.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • The original poster attempts to calculate the maximum height of the rocket using kinematic equations but does not account for the rocket's continued ascent after the motor stops. Other participants suggest considering the velocity at the moment the motor stops and calculating the additional height gained during the subsequent deceleration phase.

Discussion Status

Participants are exploring different interpretations of the rocket problem, particularly regarding the effects of the rocket's velocity after the motor stops. Some guidance has been offered regarding the need to calculate the additional height gained after the motor ceases operation.

Contextual Notes

There is a mention of the equations of motion being used, and the discussion highlights the importance of understanding the phases of motion in both problems. The second problem also introduces a quadratic equation approach, indicating a different method of analysis.

ArcadianGenesis
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Homework Statement


A rocket launched from the ground rises vertically with an acceleration of 30 m/s/s for 9.1 s until its motor stops. The acceleration of gravity is 9.8 m/s/s. Disregarding air resistance, what maximum height above the ground will the rocket achieve? Answer is units of km.


Homework Equations


v = v(initial) + at
x = x(initial) + v(initial)t + (1/2)at^2


The Attempt at a Solution


h = x - x(initial)
h = (1/2)(30)(9.1)^2
h = 1242 m
h = 1.242 km

I also tried it another way and got the same result. I don't see what could possibly be wrong.
 
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I actually have another problem of the same type - didn't want to create another topic.

Homework Statement


A motorist is traveling at 12 m/s when he sees a deer in the road 43 m ahead. If the maximum negative acceleration of the vehicle is -7 m/s/s, what is the maximum reaction time (delta)t of the motorist that will allow him to avoid hitting the deer? Answer in units of s.


Homework Equations


v = v(initial) + at
x = x(initial) + v(initial)t + (1/2)at^2


The Attempt at a Solution


43 = 0 + 12t + (1/2)(-7)t^2
0 = -43 + 12t - (7/2)t^2
using quadratic formula...
t = .457 s
 
the application of concepts is wrong...

after the motor stops the rocket still has some velocity with it to carry it upward...

calculate that distance after the motor stops and then add it to get the max height..
 
Like guru said, you forgot its still traveling at some speed. When the rocket motor stops its not yet at its highest point. 30 m/s² for 9.1 seconds. Is an average speed of (9,1*30)/2 = 136,5 m/s.
During the rocket-powered part of the flight it travels 136,5 * 9,1 = 1242,15 m.

Now it starts decelerating. Remember, it still has a lot of speed left. the speed at t=9,1 would be 9,1*30 = 273 m/s
the deceleration is 9,8 m/s². it will take about 27,857142857142857142857142857143 seconds to reach v=0. the average speed is again 136,5. 27,857142857142857142857142857143*136,5 = 3802,5 m

your now at the highest point and have traveled 3802,5 m + 1242,15 m = 5044,65 m. (5 km)
 
Last edited:

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