Answer
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Hint: Assuming the maximum velocity, the one relation that relates all the given quantity is $v = u + at$. Using this and dividing the entire time span in two parts, the question can be solved easily.
Complete step by step answer:
Let us consider the car accelerates from rest at a constant rate $\alpha$ for time $t_{\alpha}$ to reach a maximum velocity of $v_{max}$. Therefore, we may write using the equation,
$v = u + at$
where, v is the final velocity, u is the initial velocity, a is the constant acceleration and t is the time taken. In this case,
$v = v_{max}, u = 0, a = \alpha, t = t_{\alpha}$
Therefore, we get,
$v_{max} = 0 + \alpha \times t_{\alpha}$ …(I)
Again, let us consider, the car decelerates from $v_{max}$ to rest at a constant rate of $\beta$ for time $t_{\beta}$. Hence, we have,
$v = 0, u = v_{max}, a = -\beta, t = t_{\beta}$
Therefore, we get,
$0 = v_{max} - \beta \times t_{\beta}$ …(II)
Drawing the car’s journey on a graph, we get
From the above graph, AB is the car's acceleration journey and BC is the decelerated journey.
From the question we know, the entire time taken is t, where $t = t_{\alpha}+t_{\beta}$. Solving Eq. (I) and (II) for $t_{\alpha}$ and $t_{\beta}$, we get,
$t_{\alpha} = \dfrac{v_{max}}{\alpha}$, $t_{\beta} = \dfrac{v_{max}}{\beta}$
Therefore, substituting this value in the below equation, we get
$t = t_{\alpha}+t_{\beta}$
$t = \dfrac{v_{max}}{\alpha} + \dfrac{v_{max}}{\beta}$
Now taking out the common term, we get
$t = v_{max}\left(\dfrac{1}{\alpha} + \dfrac{1}{\beta}\right)$
Taking LCM and solving, we get
$t = v_{max}\left(\dfrac{\alpha+\beta}{\alpha\beta}\right)$
$v_{max} = \left(\dfrac{\alpha\beta}{\alpha+\beta}\right)t$
Therefore, the correct option is D.
Note: If you look closely into the options, you would notice that they are of different dimensions. Hence, without even doing the sum, by performing a simple dimensional analysis, one can say the correct answer is D. It’s a trick that comes handy in a lot of such sums.
Complete step by step answer:
Let us consider the car accelerates from rest at a constant rate $\alpha$ for time $t_{\alpha}$ to reach a maximum velocity of $v_{max}$. Therefore, we may write using the equation,
$v = u + at$
where, v is the final velocity, u is the initial velocity, a is the constant acceleration and t is the time taken. In this case,
$v = v_{max}, u = 0, a = \alpha, t = t_{\alpha}$
Therefore, we get,
$v_{max} = 0 + \alpha \times t_{\alpha}$ …(I)
Again, let us consider, the car decelerates from $v_{max}$ to rest at a constant rate of $\beta$ for time $t_{\beta}$. Hence, we have,
$v = 0, u = v_{max}, a = -\beta, t = t_{\beta}$
Therefore, we get,
$0 = v_{max} - \beta \times t_{\beta}$ …(II)
Drawing the car’s journey on a graph, we get
From the above graph, AB is the car's acceleration journey and BC is the decelerated journey.
From the question we know, the entire time taken is t, where $t = t_{\alpha}+t_{\beta}$. Solving Eq. (I) and (II) for $t_{\alpha}$ and $t_{\beta}$, we get,
$t_{\alpha} = \dfrac{v_{max}}{\alpha}$, $t_{\beta} = \dfrac{v_{max}}{\beta}$
Therefore, substituting this value in the below equation, we get
$t = t_{\alpha}+t_{\beta}$
$t = \dfrac{v_{max}}{\alpha} + \dfrac{v_{max}}{\beta}$
Now taking out the common term, we get
$t = v_{max}\left(\dfrac{1}{\alpha} + \dfrac{1}{\beta}\right)$
Taking LCM and solving, we get
$t = v_{max}\left(\dfrac{\alpha+\beta}{\alpha\beta}\right)$
$v_{max} = \left(\dfrac{\alpha\beta}{\alpha+\beta}\right)t$
Therefore, the correct option is D.
Note: If you look closely into the options, you would notice that they are of different dimensions. Hence, without even doing the sum, by performing a simple dimensional analysis, one can say the correct answer is D. It’s a trick that comes handy in a lot of such sums.
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