Question

Can someone help me with this...

A car traveling at 41 ft/sec decelerates at a constant 4 feet per second per second. How many feet does the car travel before coming to a complete stop?

Answer 1

i can give it a shot, but no guarantees

Answer 2

you should post this question in physics group

Answer 3

acceleration is the second derivative of the position function p, the first derivative is the speed (velocity) v
you know that
\[a(t)=-4 = v'(t) \] that means
\[v(t)=-4t+C\] and we know that the initial speed is 41 so
\[v(0)-41\implies v(t)=-4t+41\]

Answer 4

sorry last line should read
\[v(0)=41\implies v(t)=-4t+41\]

Answer 5

the car will stop when the velocity is 0 so that should be at time
\[v(t)=-4t+41=0\] meaning
\[t=\frac{41}{4}\]

Answer 6

now we want the position function, p and we know
\[p'(t)=v(t)=-4t+41\] making
\[p(t)=-2t^2+41t\]

Answer 7

correct me if i am making a colossal mistake, i don't know any physics whatsoever

Answer 8

this will be goin to b quadritic one

Answer 9

no , i think now you simply replace t by 41/4 to get the answer

Answer 10

v=0, u=-41t

Answer 11

Uhm, I replaced t in the equation but that answer was wrong... I don't know.

Answer 12

sorry dint read full question

Answer 13

if i am not mistaken now it is a matter of computing
\[p(\frac{41}{4})\] so finish. i get 200.125

Answer 14

actually i get 210.125

Answer 15

answer is 10.25

Answer 16

then i am wrong

Answer 17

It was 210.125 thanks everyone ^-^

Answer 18

well, i am not so sure about the 10.25 where did that come from? it seem unlikely to me

Answer 19

lolx i was doin programmin so did it wrong lolx