A driver in a car traveling at a speed of
75 km/h sees a cat 125 m away on the road.
How long will it take for the car to accelerate
uniformly to a stop in exactly 122 m?
Answer in units of s.

- Anikate

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- chestercat

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- Anikate

any clue??

- anonymous

@pooja195 Help me

- anonymous

@pooja195 im new to helping

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## More answers

- Anikate

do u know the answer??

- Anikate

@DylanisNOW u know the answer??

- anonymous

Im trying to get @pooja195

- Anikate

i thuink she offline, but do know the answer to this question??

- Anikate

she appeears to be offline??

- Anikate

yea she offline

- anonymous

Hold on...Im trying to figure it out. Please dont rush!

- Anikate

ok

- anonymous

@pooja195 No she isnt

- Anikate

ok, do u know how to do it though?

- anonymous

Give me a sec please

- Anikate

ok np

- Anikate

@pgpilot326 do u know how to do this? its a physics question

- anonymous

Think of the units.

- Anikate

ok now?

- anonymous

units are 75km/h right?

- Anikate

yes

- anonymous

Wait sorry hold on thats not it thats only 1

- Anikate

ok

- anonymous

75km/h / 125 m=?

- Anikate

600m/h

- anonymous

Okay do you have answer choices?

- Anikate

nopee

- Anikate

do u havea roungh estimate of what the andswer could be??

- anonymous

Think of it like this. Every second the speed decreases by?

- anonymous

We cant give answers...Its just help to get you to understand it!

- anonymous

Think of it like this. Every second the speed decreases by?

- Anikate

the speed decreases by seconds(squared)

- anonymous

Ok that is?

- Anikate

idk.....

- anonymous

.... @pgpilot326 can you help me help him?

- anonymous

Anyways..

- Anikate

dylanis, can u just post a step by step explaination for it

- anonymous

Ok

- anonymous

Step 1:units are 75km/h

- anonymous

Step 2:75km/h / 125=600m/h

- anonymous

Okay now we are at step 3

- anonymous

if a is a constant, integrate to get velocity and again to get position.
the velocity should be linear and the position a quadratic.

- anonymous

since you know the initial and final speed you should be able to calc. the time

- anonymous

pgpilot should we make a equation?

- anonymous

should make 2 equations... velocity and position. you'll need both.

- Anikate

can you please show bothh?

- Anikate

and solve with steps please?

- anonymous

\[v \left( t \right)=v_0+at\]and\[x \left( t \right)=x_0+v_0t+\frac{ 1 }{ 2 }at^2\]

- anonymous

pg you are right good job btw :)

- anonymous

at t=0, initial speed (\(v_0\)) is 75 km/h and initial position is 0 m.
at t=\(t_1\), final speed (\(v_f\)) is 0 km/h and final position is 122 m.

- anonymous

you need to solve for \(t_1\)

- Anikate

ok

- Anikate

how so??

- Anikate

which equation am i putting all this in??

- anonymous

phone, just a minute

- anonymous

\[v \left( t_1 \right)=0=75 +a\cdot t_1\]
\[x \left( t_1 \right)=122=0+75t_1+\frac{ 1 }{ 2 }at_1^2\]

- anonymous

find roots of quadratic in terms of a.

- Anikate

u can handwrite it if u wish, ill prob be able to decode it

- Anikate

|dw:1443142704493:dw|

- Anikate

isnt htat what we have now?? what now??

- Anikate

can u show rest of steps

- anonymous

solve velocity eqn for t1 and position for t1 and equate

- anonymous

to solve for a. once you have a you can find t1

- Anikate

how do u solve for t1?? there are 2 variables in the equation itself

- anonymous

from velocity:
t1 = -75/a

- anonymous

use quadratic to solve the position:
\[0=-122+75t_1+\frac{ 1 }{ 2 }at_1^2\]for t1 in terms of a

- Anikate

|dw:1443143049231:dw|

- Anikate

|dw:1443143204081:dw| thats as far as i can get

- Anikate

can u please explain how u would do it, and go past that step

- anonymous

|dw:1443143218606:dw|

- anonymous

use the quadratic formula

- anonymous

|dw:1443143430909:dw|then solve for a

- Anikate

cant root a variable

- Anikate

keep going

- anonymous

you'll have to square to get rid of square root

- Anikate

ok then, can u plz post work till answer?

- Anikate

kinda hard to follow step by step

- anonymous

k but it will take a minute

- Anikate

take ur time

- anonymous

\[122=75\left( \frac{ -75 }{ a } \right)+\frac{ a\left( -75 \right)^2 }{ 2a^2} \Rightarrow 122 =-\frac{ 5625 }{ a } +\frac{ 5625 }{ 2a }\Rightarrow 244a = -11250+5625\]

- anonymous

\[a=-\frac{ 5625 }{244 } \Rightarrow t_1 = \frac{ -75 }{ -\frac{ 5625 }{ 244 } }=\frac{ 244 }{ 75} \text{ sec}\]

- Anikate

AAH ok! I see! thanks a lot! @pgpilot326

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