Question

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

Answer 1

Distance Travelled in nth Second

\[Dn = u + \frac{ a }{ 2 }[2n-1]\]

Since the Question Talks about retardation

\[Dn = u - \frac{ a }{ 2 }[2n-1]\]

Answer 2

\[v = u -at\]

\[u = at\]

t - 1 = n

t = n +1

u = a (n+1)

Answer 3

Nw Substitute the Value of u in the 1st equation

Answer 4

\[Dn = an + a - an + a/2 = 3a/2\]

Answer 5

Thank you! (:

Answer 6

Welcome...

Answer 7

why is t-1=n?

Answer 8

secondlast second of its motion

Answer 9

let t be the total time

Answer 10

secondlast = t-1

Answer 11

oh.. haha.. ty again!