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