Question

A car is moving in a straight line with speed $18km{{h}^{-1}}$, it is stopped in 5s by applying the brakes. Find:
(i) the initial speed of the car in $m{{s}^{-1}}$
(ii) the retardation and
(iii) the speed of the car after 2s of applying brakes.
A. $5m{{s}^{-1}},1m{{s}^{-2}},2m{{s}^{-1}}$
B. $5m{{s}^{-1}},2m{{s}^{-2}},2m{{s}^{-1}}$
C. $2m{{s}^{-1}},1m{{s}^{-2}},3m{{s}^{-1}}$
D. None of these

Answer 1

Hint: Since 1 km is 1000m and 1 hr is 60 min i.e. 3600 seconds, putting these values will give us the desired result. To calculate the retardation of the car use the formula for constant acceleration of a body. Then use the same formula to find the speed after 2s of applying brakes.

Formula used:
$a=\dfrac{v-u}{t}$
Where $a$ is the acceleration, $v$ is the final velocity, $u$ is the initial velocity and $t$ is the time.

Complete step by step answer:
(i) Here the initial speed of the car is given to be $18km{{h}^{-1}}$. We are supposed to cover this speed into $m{{s}^{-1}}$.
Let the initial velocity of the car be u.
Then, this means that $u=18km{{h}^{-1}}$.
We know that $1km=1000m$ and $1h=3600s$.
By substituting these values we get that $u=18\times \dfrac{1000m}{3600s}=5m{{s}^{-1}}$.
This means that the initial speed of the car is $5m{{s}^{-1}}$.

(ii) Let us assume that the rate at which the velocity of the car decreases to zero is constant. Then this means that the retardation (negative acceleration) of the car is constant.
For uniform acceleration, $a=\dfrac{v-u}{t}$ ….. (i), where a is acceleration, v is the velocity of the car at time t and u is the initial velocity of the car.
In this case, $u=5m{{s}^{-1}}$and at time $t=5s$, the car comes to rest. Then this means that $v=0m{{s}^{-1}}$.
Substitute these values in (i).
$\therefore a=\dfrac{0-5}{5}=-1m{{s}^{-2}}$.
This means that the car is accelerating in the opposite direction at a rate of $1m{{s}^{-2}}$. In other words, its retardation is $1m{{s}^{-2}}$.

(iii) Let at time $t=2s$, the velocity of the car be v’.
Substitute $u=5m{{s}^{-1}}$, $a=-1m{{s}^{-2}}$, $t=2s$ and $v=v'$.
$\Rightarrow -1=\dfrac{0-v'}{2}$
$\therefore v'=2m{{s}^{-1}}$.
Therefore, the speed of the car after 2s of applying brakes is $2m{{s}^{-1}}$.

Hence, the correct option is A.

Note: For conversion of units from $km{{h}^{-1}}$ to $m{{s}^{-1}}$ one can also remember a simple conversion factor of 5/18. When 5/18 is multiplied to any speed in $km{{h}^{-1}}$, the result that we get is in $m{{s}^{-1}}$.Similarly, to convert unit from $m{{s}^{-1}}$ to $km{{h}^{-1}}$ we can remember the inverse of this conversion factor i.e. 18/5. When 18/5 is multiplied to a speed in ms-1, we get the result in $km{{h}^{-1}}$.