Motion with constant acceleration

In summary, the rocket will achieve a maximum height of approximately 5 km above the ground before reaching its highest point. This is calculated by taking into account the initial acceleration of 30 m/s/s for 9.1 seconds, followed by deceleration of 9.8 m/s/s for 27.857 seconds. The total distance traveled is 5044.65 m or 5 km.
  • #1
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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  • #2
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
 
  • #3
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..
 
  • #4
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)
 
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FAQ: Motion with constant acceleration

What is constant acceleration?

Constant acceleration refers to the rate of change of an object's velocity remaining constant over time. This means that the object's speed increases by the same amount in each unit of time.

How is constant acceleration different from uniform motion?

Uniform motion is when an object moves at a constant speed in a straight line, while constant acceleration involves a change in velocity over time. In constant acceleration, the object's speed will increase or decrease at a constant rate, while in uniform motion, the object's speed remains the same.

What is the formula for calculating constant acceleration?

The formula for calculating constant acceleration is a = (v - u) / t, where a is the acceleration, v is the final velocity, u is the initial velocity, and t is the time elapsed.

What is the difference between positive and negative constant acceleration?

Positive constant acceleration means that the object's velocity is increasing over time, while negative constant acceleration means that the object's velocity is decreasing over time. This can also be referred to as acceleration in the positive or negative direction.

Can an object have constant acceleration and changing velocity?

Yes, an object can have constant acceleration and changing velocity. This is because acceleration is the rate of change of velocity, so even if the acceleration remains constant, the velocity can still change if the object's initial velocity or time changes.

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