- #1

chwala

Gold Member

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- Homework Statement
- See attached

- Relevant Equations
- kinematics

Wawawawawawa! This was a tough one...

Find the question below;

Find my approach below:

We have the following equations;

1. ##v=30-0.5t##

2. ##v=30-1.5t##

Now the car changes its acceleration at some point i.e from ##-\frac{1}{2}## ##m/s^2## to ##-\frac{3}{2}## ##m/s^2## ...

I considered the following graphs for the two equations, assuming that the second equation was also starting from the point ##v=30####m/s##

Now the graph of ##v=30-1.5t## will give us some perspective on how it relates to the graph of ##v=30-0.5t## at some point...noting that the desired graph will be parallel to ##v=30-1.5t##.

Now i attempted to find the scale factor ##m## connecting the two functions by letting,

##-\frac{3}{2}##m = ##-\frac{1}{2}##

##→m##= ##\frac{1}{3}## therefore plotting the graph,

##v##= ##\frac{1}{3}t## realizes the point that we were looking for where, ##(t,v)##=##(36,12)##

Therefore, the speed of the car at the point where deceleration is increased is ##12## m/s and the time taken for the car to stop is given by

using ##v=u +at##

##0 ##= ##12##+ ##-\frac{3}{2}####t##

##t=8##, therefore time taken for car to stop is ##36+8=44##seconds

bingo guys

There may be a better approach to what i have done...

Find the question below;

Find my approach below:

We have the following equations;

1. ##v=30-0.5t##

2. ##v=30-1.5t##

Now the car changes its acceleration at some point i.e from ##-\frac{1}{2}## ##m/s^2## to ##-\frac{3}{2}## ##m/s^2## ...

I considered the following graphs for the two equations, assuming that the second equation was also starting from the point ##v=30####m/s##

Now the graph of ##v=30-1.5t## will give us some perspective on how it relates to the graph of ##v=30-0.5t## at some point...noting that the desired graph will be parallel to ##v=30-1.5t##.

Now i attempted to find the scale factor ##m## connecting the two functions by letting,

##-\frac{3}{2}##m = ##-\frac{1}{2}##

##→m##= ##\frac{1}{3}## therefore plotting the graph,

##v##= ##\frac{1}{3}t## realizes the point that we were looking for where, ##(t,v)##=##(36,12)##

Therefore, the speed of the car at the point where deceleration is increased is ##12## m/s and the time taken for the car to stop is given by

using ##v=u +at##

##0 ##= ##12##+ ##-\frac{3}{2}####t##

##t=8##, therefore time taken for car to stop is ##36+8=44##seconds

bingo guys

There may be a better approach to what i have done...

Last edited: