## yrelhan4 3 years ago The driver of a car moving on a straight road applies brakes to come to rest, with a constant retardation 'a'. Assuming that the time of motion is more than 2 s, the distance covered by the car in secondlast second of its motion is

1. anonymous

Distance Travelled in nth Second $Dn = u + \frac{ a }{ 2 }[2n-1]$ Since the Question Talks about retardation $Dn = u - \frac{ a }{ 2 }[2n-1]$

2. anonymous

$v = u -at$ $u = at$ t - 1 = n t = n +1 u = a (n+1)

3. anonymous

Nw Substitute the Value of u in the 1st equation

4. anonymous

$Dn = an + a - an + a/2 = 3a/2$

5. yrelhan4

Thank you! (:

6. anonymous

Welcome...

7. yrelhan4

why is t-1=n?

8. anonymous

secondlast second of its motion

9. anonymous

let t be the total time

10. anonymous

secondlast = t-1

11. yrelhan4

oh.. haha.. ty again!