Initial velocity problem.

C

calculusdunce

New member
Joined
Jun 11, 2021
Messages
6
Here's the problem:
62_08.png

I've been struggling with this kind of problem, so I thought I'd get help.

I know that the formula to use here is [MATH]\int a(t)dt = v(t)+c[/MATH]. I think.

And I can see that [MATH]v(6)=10[/MATH]. Where do I go from here?
 
Since these are all straight lines, I wouldn't bother with integration- that's just the area under the line. Assuming each horizontal step is one second and each vertical step is one m/s^2, the area under the line from t= 0 to 4 is the area of the trapezoid with "bases" of 4 m/s^2 and 7 m/s^2 and "height" 4 s. The increase in velocity is ((7+ 4)/2)(4)= 11*2= 22 m/s. If the initial velocity was "v" then the speed after 4 seconds is v+ 22 m/s. From 4 seconds to 6 seconds, we have a horizontal line at a= 7. The area of a rectangle with height 7 and length 3 is 3(7)= 21. The speed after 6 seconds is v+ 22+ 21= 10. v= 10- 22- 21= -33 m/s.
 
[math]\displaystyle v(6) - v(0) = \int_0^6 a(t) \, dt[/math]
[math]10 - v(0) = 36[/math]
 
HallsofIvy said:
Since these are all straight lines, I wouldn't bother with integration- that's just the area under the line. Assuming each horizontal step is one second and each vertical step is one m/s^2, the area under the line from t= 0 to 4 is the area of the trapezoid with "bases" of 4 m/s^2 and 7 m/s^2 and "height" 4 s. The increase in velocity is ((7+ 4)/2)(4)= 11*2= 22 m/s. If the initial velocity was "v" then the speed after 4 seconds is v+ 22 m/s. From 4 seconds to 6 seconds, we have a horizontal line at a= 7. The area of a rectangle with height 7 and length 3 is 3(7)= 21. The speed after 6 seconds is v+ 22+ 21= 10. v= 10- 22- 21= -33 m/s.
Click to expand...
Professor Halls! If the base goes from 4 to 6 that is not a distance of 3! From 4 to 7 is a distance of 3.
 
calculusdunce said:
Could I have more help? Sorry...
Click to expand...
You are given a graph. Horizontal axis is the time axis. What does each unit on horizontal axis display - 10 secs .... 50 mins....100 hours...
And What does each unit on vertical axis display?
1623449432001.png
 
skeeter said:
[math]\displaystyle v(6) - v(0) = \int_0^6 a(t) \, dt[/math]
[math]10 - v(0) = 36[/math]
Click to expand...
I think that this is a perfect answer.

You don't need to complicate things. Area under the curve [MATH]= \Delta v[/MATH] (Change in velocity)

Calculate the area, and then Final Velocity [MATH]-[/MATH] Initial Velocity [MATH]= [/math] Area

skeeter gave you a beautiful answer.
 
Subhotosh Khan said:
How are you calculating area when the unit magnitude of axes are unknown?
Click to expand...

Well, I guess I assumed that the units on the horizontal axis were seconds & meters per second squared on the vertical.

suffice it to say [MATH]v(6) - v(0) = \int_0^6 a(t) \, dt[/MATH] is correct w/o any assumptions.
 
skeeter said:
Well, I guess I assumed that the units on the horizontal axis were seconds & meters per second squared on the vertical.

suffice it to say [MATH]v(6) - v(0) = \int_0^6 a(t) \, dt[/MATH] is correct w/o any assumptions.
Click to expand...
I was more concerned with the "magnitude" (scale) - opposed to units.

i.e. 1 unit on x-axis =1 sec or 5 secs.
 
You must log in or register to reply here.
Top