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# A car moving with the speed of** 108 km/h** on a straight road, decreases its speed uniformly to **90 km/h** in **5 seconds**. The displacement of the car in the last second of its motion, before it comes to a halt?

**Given:**

Initial velocity of the car, $u=108km/h=\frac {5}{18}\times 108=30m/s$ (converted km/h to m/s)

Final velocity of the car, $v=90km/h=\frac {5}{18}\times 90=25m/s$ (converted km/h to m/s)

Time taken, $t$ = 5s

**To find: **Displacement, $(s)$ of the car in the last second of its motion, before it comes to halt.

**Solution:**

For finding acceleration $(a)$ we use the first equation of motion-

$v=u+at$

Putting the given values, we get-

$25=30+a(5)$

$25-30=5a$

$a=\frac {-5}{5}$

$a=-1m/s^2$

Thus, the acceleration of the car is $-1m/s^2$.

Now, we need to calculate the time taken $(t')$ by the car to come to rest, so the final velocity will be zero $(0)$.

Again, putting the values in the formula of first law of motion we get-

$0=30+(-1)t'$

$-30=-1t'$

$t'=-30s$

Thus, the car comes to rest in $30\ seconds$.

To find the displacement of the car in the 30th [last] second-

The formula for the distance covered by a body in $nth\ second$, if $u$ is the initial velocity and $a$ is the uniform acceleration of the object is given by-

$S_{nth}=u+\frac {a}{2}(2n-1)$

Putting the given values, we get-

$S_{30}=30+\frac {(-1)}{2}(2\times {30}-1)$

$S_{30}=30-\frac {1}{2}(59)$

$S_{30}=30-\frac {1}{2}\times (59)$

$S_{30}=30-29.5$

$S_{30}=0.5m$

Thus, the displacement of the car in the last second of its motion is **0.5m.**

**
**

**Note:**

To find the distances, we use Newton's second law of motion which states that-

$s=ut+\frac {1}{2}at^2$

Now, to find the distance covered in $nth\ second$, if $u$ is the initial velocity and $a$ is the uniform acceleration of the object, then this second formula is expanded by putting $t=n-1$ and $n$ respectively. And, its subtraction gives the difference between the distances covered in $n$ and $n-1$ seconds, which is nothing but the distance covered in the $n_th$ second.

First, putting the value of $t=n$ in the second law of motion we get-

$s=un+\frac {1}{2}an^2$ ----------------------(i)

Second, putting the value of $t=n-1$ in the second law of motion we get-

$s=u(n-1)+\frac {1}{2}a(n-1)^2$ ----------------------(ii)

Subtracting $(ii)$ from $(i)$ we get-

$S_{nth}=[u.n+\frac {1}{2}an^2]-[u(n-1)+\frac {1}{2}a(n-1)^2]$

$S_{nth}=u+\frac {a}{2}(2n-1)$