How do I find intial acceleration?

In summary, the distance covered by each crate is 20m and they start at rest. Using the formula v2 = u2 + 2as, with u = 0ms-1, s = 20m, and a = 2.4148ms-2, we can calculate the final speed of each crate to be either v2 = 9.8281 ms-1 or v = 96.592 ms-1.
  • #1
vpa021
11
0
You accelerate at a m/s2 for 4.0 s, then coast for 8.0 seconds, then
decelerate to a standstill in 4.0 s. In total, you travel 144 meters. What was your
initial acceleration?

Please show your work as i need to learn from you.
 
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  • #2
Well, we don't usually do the work for the students here but we try to nudge them in the right direction.

The right direction here is how would you write up an equation for the distance traveled as a function of the acceleration a (what is the deceleration as a function of a? can you tell that by the numbers given?).
 
  • #3
this is what i did ΔX = v_0t+.5at^2

Given the total Traveled ΔX= 144m v_0=0 t=4s

My equation looks something like this

144m=0+(4s)+.5a(4s)^2

144m=8s^2a

a=18m/s^2

but i think this is wrong because i took the total (144m) traveled. I don't know how to get the acceleration at 0m and 4s.
 
  • #4
Ok, that's a start.

But your equation assumes that the whole 144m are traveled in the initial 4 seconds which is not true, the 144 m are traveled in 16 seconds but with varying speed (except between 4 and 12 seconds) so you have to break the distance down into three parts and each part gets its own term.

Can you see how you get the velocity at 4 seconds (and thus at 12 seconds also) as a function of a?
 
  • #5
ojs said:
Ok, that's a start.

But your equation assumes that the whole 144m are traveled in the initial 4 seconds which is not true, the 144 m are traveled in 16 seconds but with varying speed (except between 4 and 12 seconds) so you have to break the distance down into three parts and each part gets its own term.

Can you see how you get the velocity at 4 seconds (and thus at 12 seconds also) as a function of a?

I surely don't understand how to break the distance into 3 parts. Unless of course to divide:

144/3 = 48 m?
 
  • #6
Ok then, what you have to do here is write up 3 terms that all show the distance traveled as a function of a.

The first 4 seconds the distance traveled is 0.5at^2, then you have to add a second term that shows the distance traveled during the next 8 seconds as a function of a and lastly a term that describes the distance traveled during the last 4 seconds.

so it will be a function like this one here:

144m = 0.5a*(t1)^2 + v1*t2 + (v1_0*t3 + 0.5a3*(t3)^2)

where a is our initial acceleration, t1 is the initial 4 seconds, v1 is the speed after 4 seconds, t2 is the 8 second interval and the last part in the parenthesis is the distance traveled during the last 4 seconds denoted by t3, v1_0 is the initial velocity of that part (how is that related to v1?) and a3 is the deceleration (how is that related to a?).

Can you continue from here?
 
  • #7
Ok then, what you have to do here is write up 3 terms that all show the distance traveled as a function of a.

The first 4 seconds the distance traveled is 0.5at^2, then you have to add a second term that shows the distance traveled during the next 8 seconds as a function of a and lastly a term that describes the distance traveled during the last 4 seconds.

so it will be a function like this one here:

144m = 0.5a*(t1)^2 + v1*t2 + (v1_0*t3 + 0.5a3*(t3)^2)

where a is our initial acceleration, t1 is the initial 4 seconds, v1 is the speed after 4 seconds, t2 is the 8 second interval and the last part in the parenthesis is the distance traveled during the last 4 seconds denoted by t3, v1_0 is the initial velocity of that part (how is that related to v1?) and a3 is the deceleration (how is that related to a?).

Can you continue from here?
 
  • #8
What formula is that?
 
  • #9
That is the well known formula

x = x_0 + v_0*t + 0.5*a*t^2

used in all three parts of the journey.

The first term in my first equation is the last part of this formula, that describes the distance traveled during the first 4 seconds.
The second term in my first equation is the second part of this formula, it describes the distance traveled during the 8 seconds that object has constant speed.
The third term in my first equation (the one inside the parenthesis) is the distance traveled during the last 4 seconds with v1 being the initial speed when it starts to decelerate and a3 being the deceleration.
 
  • #10
That is the well known formula

x = x_0 + v_0*t + 0.5*a*t^2

used in all three parts of the journey.

The first term in my first equation is the last part of this formula, that describes the distance traveled during the first 4 seconds.
The second term in my first equation is the second part of this formula, it describes the distance traveled during the 8 seconds that object has constant speed.
The third term in my first equation (the one inside the parenthesis) is the distance traveled during the last 4 seconds with v1 being the initial speed when it starts to decelerate and a3 being the deceleration.
 
  • #11
Thank you ojs.

my professor never gave us this formula. Hence, i have a problem with this question and similar questions as this.
 
  • #12
Hi guys, just wondering if anyone could help me out?
 
  • #13
wilko2008 said:
Hi guys, just wondering if anyone could help me out?

The distance covered by each crate is 20m. Assuming that each crate starts at rest, calculate the speed of each crate at the end of the 20m journey.

I have used v2 = u2 + 2as

u = start speed = 0ms-1
s = distance traveled = 20m
a = acceleration = 2.4148ms-2
v2 = final speed ms-1

how would i show the final speed? v2 = 9.8281 ms-1 or
v = 96.592 ms-1
 

1. How do I calculate initial acceleration?

The formula for calculating initial acceleration is a = (vf - vi)/t, where a is the acceleration, vf is the final velocity, vi is the initial velocity, and t is the time interval.

2. What is the difference between initial acceleration and average acceleration?

Initial acceleration refers to the acceleration at the beginning of a motion, while average acceleration refers to the overall acceleration during a certain period of time.

3. Can initial acceleration be negative?

Yes, initial acceleration can be negative if the initial velocity is greater than the final velocity, indicating a decrease in speed.

4. How is initial acceleration related to velocity and time?

Initial acceleration is directly proportional to the change in velocity and inversely proportional to the time interval. This means that a larger change in velocity or a shorter time interval will result in a higher initial acceleration.

5. Why is initial acceleration important in motion?

Initial acceleration is important because it helps us understand the rate at which an object's velocity changes at the beginning of its motion. It is also used to analyze the motion of objects and determine the forces acting upon them.

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