Question

An engine of a vehicle can produce a maximum acceleration of $4\,{\text{m}} \cdot {{\text{s}}^{ - 2}}$. Its breaks can produce a maximum retardation of $6\,{\text{m}} \cdot {{\text{s}}^{ - 2}}$. The minimum time at which it can cover a distance of $3\,{\text{km}}$ is:

A. $30\,{\text{s}}$

B. $40\,{\text{s}}$

C. $50\,{\text{s}}$

D. $60\,{\text{s}}$

Answer 1

Use the formula for the displacement of an object. This formula should give the relation between the maximum acceleration of the object, maximum retardation of the object, displacement of the object and the minimum time required for the displacement. Convert the unit of the displacement of the vehicle in the SI system of units i.e. from kilometer to meter and then use it in the formula.

Formula used:

The displacement $S$ of an object is given by

$S = \dfrac{{\alpha \beta }}{{2\left( {\alpha + \beta } \right)}}{t^2}$ …… (1)

Here, $\alpha$ is the maximum acceleration of the object, $\beta$ is the maximum retardation of the object and $t$ is the minimum time for the displacement of the object.

Complete step by step answer:

The maximum acceleration of the vehicle engine is $4\,{\text{m}} \cdot {{\text{s}}^{ - 2}}$ and the minimum retardation of the vehicle engine is $6\,{\text{m}} \cdot {{\text{s}}^{ - 2}}$.

$\alpha = 4\,{\text{m}} \cdot {{\text{s}}^{ - 2}}$

$\beta = 6\,{\text{m}} \cdot {{\text{s}}^{ - 2}}$

Let the vehicle engine cover a distance of $3\,{\text{km}}$ in time $t$.

$S = 3\,{\text{km}}$

Convert the unit of the displacement of the vehicle engine in the SI system of units.

$S = \left( {3\,{\text{km}}} \right)\left( {\dfrac{{{{10}^3}\,{\text{m}}}}{{1\,{\text{km}}}}} \right)$

$\Rightarrow S = 3000\,{\text{m}}$

Hence, the displacement of the vehicle is $3000\,{\text{m}}$.

Determine the time in which the vehicle covers a distance of $3000\,{\text{m}}$.

Rewrite equation (1).

$S = \dfrac{{\alpha \beta }}{{2\left( {\alpha + \beta } \right)}}{t^2}$

Rearrange the above equation for time $t$.

$t = \sqrt {\dfrac{{2S\left( {\alpha + \beta } \right)}}{{\alpha \beta }}}$

Substitute $3000\,{\text{m}}$ for $S$, $4\,{\text{m}} \cdot {{\text{s}}^{ - 2}}$ for $\alpha$ and $6\,{\text{m}} \cdot {{\text{s}}^{ - 2}}$ for $\beta$ in the above equation.

$t = \sqrt {\dfrac{{2\left( {3000\,{\text{m}}} \right)\left( {\left( {4\,{\text{m}} \cdot {{\text{s}}^{ - 2}}} \right) + \left( {6\,{\text{m}} \cdot {{\text{s}}^{ - 2}}} \right)} \right)}}{{\left( {4\,{\text{m}} \cdot {{\text{s}}^{ - 2}}} \right)\left( {6\,{\text{m}} \cdot {{\text{s}}^{ - 2}}} \right)}}}$

$\Rightarrow t = \sqrt {\dfrac{{60000}}{{24}}}$

$\Rightarrow t = \sqrt {2500}$

$\therefore t = 50\,{\text{s}}$

Therefore, the time required for the vehicle engine to cover a distance of $3\,{\text{km}}$ is $50\,{\text{s}}$.

**So, the correct answer is “Option C”.
**
Note:

Convert the unit of the displacement from kilometer to meter. In order to get the correct answer, substitute the values of all the physical quantities in the formula in the same SI system of units including

displacement of the object and the minimum time required for the displacement in the formula used in the above question.