a car is moving on a straight road with a speed of 20m/s. At t=0 the driver applies brakes after watching an obstacle 150m ahead. after application of brakes the car retards with 2m/s^2. Find the position of the car from the obstacle at t=15

Question

a car is moving on a straight road with a speed of 20m/s. At t=0 the driver applies brakes after watching an obstacle 150m ahead. after application of brakes the car retards with 2m/s^2. Find the position of the car from the obstacle  at t=15

anonymous · Answer

i got the answer
 do u want to answer?

anonymous · Answer

still reading the question..

stormfire1 · Answer

$$d=V_it+\frac{1}{2}at^2$$$$d=20 m/s(15)+\frac{1}{2}(-2m/s^2)(15)^2$$$$d=75m$$So the car is still 75m from the obstacle after 15 seconds.

stormfire1 · Answer

Is that what you got?

anonymous · Answer

No. I was asking cause i was getting the wrong answer and doing the same mistake.
instead of 2nd equation use 3rd equation of motion then u will ge the answer.

stormfire1 · Answer

yea...I was looking at it and it doesn't seem right at all...let me look again

anonymous · Answer

the car will stop after 10 seconds of breaking right ? ?

stormfire1 · Answer

yea

stormfire1 · Answer

so yea...I shoulda used 10s

anonymous · Answer

no 15

stormfire1 · Answer

Well, the car isn't going to start going backwards from braking :P

anonymous · Answer

after the 1st second the speed will be 18 after second 16 and so on

anonymous · Answer

use 3rd equation of motion
$$v^2-u^2=2as$$

anonymous · Answer

Initial Velocity (u)=20m/s
Acceleration (a)  =-2m/s^2
Final velocity v =0
Time(t)=15s
#rd equation of motion
$$v^2-u^2=2as$$
$$0^2-(-20)^2]=2*(-2)s$$
$$-400=-4s$$
$$-400/-4=s$$
$$100=s$$

anonymous · Answer

3rd equation*

stormfire1 · Answer

Yea, you get 100s when you use the first equation with 10s also :)

stormfire1 · Answer

I just didn't consider that I had to drop off the last 5 seconds since the car was stopped by then.

anonymous · Answer

yeah but we can't change the given values

anonymous · Answer

u r also right but the values are given

stormfire1 · Answer

You're not changing anything...it's part of solving the problem in my opinion.  If the question had stated 2000seconds...the reality is that the car is still stopped after 10s.

anonymous · Answer

exactly

stormfire1 · Answer

Besides, when you did your equation, you *assumed* Vf was 0 :)

anonymous · Answer

u are also right man. consider this <======

anonymous · Answer

its not assumption if the car stops it means v= 0

stormfire1 · Answer

Did the question state that the car stopped?

stormfire1 · Answer

No :)

anonymous · Answer

yes

stormfire1 · Answer

What part?  It says it's slowing down at 2 m/s...

anonymous · Answer

no sorry :p

stormfire1 · Answer

So YOU CHANGED THE PROBLEM!!  just kidding

anonymous · Answer

lol

stormfire1 · Answer

Here...let's graph it too so we can beat this problem to death :)