Question

The engine of a futuristic number. Powered cars for which power and speed can have (fantastic values kph) can produce a maximum acceleration of $5m/{s^2}$ and its brakes can produce a maximum retardation of $10m/{s^2}$. The minimum time, in which a person can reach his workplace, located 1.5 km away from his home using this car is.
A. 5 sec
B. 10 sec
C. 15 sec
D. 30 sec

Answer 1

In this type of question, we know about the acceleration and retardation and distance covered by the car then you will use the formula of distance related to acceleration and retardation. $S = \dfrac{1}{2}\dfrac{{\alpha \beta {t^2}}}{{\left( {\alpha + \beta } \right)}}$
where, $\alpha$ is acceleration.
$\beta$ is retardation.
“s” is distance covered.
“t” is time in seconds.

Complete step by step answer:
Acceleration is defined as the change in velocity per unit time. Acceleration is measured in $m/{s^2}$. The negative acceleration is called retardation or deceleration. Its unit is the same as acceleration.
According to question:
Given,
Distance, S = 1.5 km = 1500 m
Acceleration F = $5m/{s^2}$
Deceleration F = $10m/{s^2}$
Time, t = ?
Now, we know that
$S = \dfrac{1}{2}\dfrac{{\alpha \beta {t^2}}}{{\left( {\alpha + \beta } \right)}}$
Where,
If a body starts from rest with acceleration $\alpha$ and then retard with retardation $\beta$. The total time taken for this journey is t and distance covered is s then,
Now, putting these given value
$1500 = \dfrac{1}{2} \times \dfrac{{5 \times 10 \times t}}{{\left( {5 + 10} \right)}}$
$1500 = \dfrac{{25 \times {t^2}}}{{15}}$
${t^2} = \dfrac{{15 \times 1500}}{{25}}$
${t^2} = 900\;\sec$
$t = \sqrt {900} \;\sec$
$t = \sqrt {30 \times 30} \;\sec$
T = 30 sec.

So, the correct answer is “Option D”.

Note: In this type of question, first we know about acceleration of a body and retardation of a body in which the velocity of the body may either increase or decrease. The change in velocity is known as acceleration. If the velocity of the body increases, acceleration is said to be positive. ... Retardation is acceleration with a negative sign, and knows about the distance formula at constant acceleration as $S = \dfrac{1}{2}\dfrac{{\alpha \beta {t^2}}}{{\left( {\alpha + \beta } \right)}}$ then, you will proceed to solve it.