
A truck has to carry a load in the shortest time from one station to another station situated at a distance L from the first. It can start up or slow down at the same acceleration or deceleration a. What maximum velocity must the truck attain to satisfy this condition?
A. $\sqrt{2La}$
B. $\sqrt{5La}$
C. $\sqrt{La}$
D. $\sqrt{3La}$
Answer
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Hint:Here we have to find the maximum velocity attained by a truck that satisfies the given condition. Condition is that the lorry should only take the shortest time to reach another station and the lorry can accelerate or decelerate. We have to find the derivative of velocity equating it to zero to find the maximum velocity.
Formula used:
Newton’s equations of motion are used.
1.$v=u+at$
where v= final velocity, u=initial velocity, a= acceleration and t= time
2.$s=ut+\dfrac{1}{2}a{{t}^{2}}$
where t= time, s= distance travelled, u=initial velocity and a= acceleration
3.${{v}^{2}}={{u}^{2}}+2as$
where v= final velocity, s= distance travelled, u=initial velocity and a= acceleration
Complete step by step solution:
Here we consider v as the maximum velocity attained by the lorry to satisfy the condition and t is the minimum time taken by the lorry to reach from one station to another. Let t’ be the duration of acceleration and t” be the time of deceleration. From newton’s first equation of motion, we have velocity,
$v=0+at'$
From the question we can write that the total distance travelled by the lorry is:
$L=\dfrac{1}{2}at'+v(t-2t')+\dfrac{1}{2}at'$
$L=\dfrac{{{v}^{2}}}{a}+vt-\dfrac{2{{v}^{2}}}{a}=vt-\dfrac{{{v}^{2}}}{a}$
From this equation we can write an equation for time taken as:
Total time taken,
$t=\dfrac{L}{v}+\dfrac{v}{a}$
For finding the maximum velocity attained differentiating the above equation with respect to velocity. That is;
$\dfrac{dt}{dv}=-\dfrac{L}{{{v}^{2}}}+\dfrac{1}{a}$
We know that when time is minimum,
$\dfrac{dt}{dv}=0$
On applying this condition, we get maximum velocity as,
$-\dfrac{L}{{v_{\max}}^2}+\dfrac{1}{a} =0 $
$\therefore {{v}_{\max }}=\sqrt{La}$
Therefore, the answer is option C.
Notes: Here we have to remember that the conditions are given. So, we have to formulate such an equation that satisfies every condition. Then only we can derive it and equate it to zero to find the extreme condition.
Formula used:
Newton’s equations of motion are used.
1.$v=u+at$
where v= final velocity, u=initial velocity, a= acceleration and t= time
2.$s=ut+\dfrac{1}{2}a{{t}^{2}}$
where t= time, s= distance travelled, u=initial velocity and a= acceleration
3.${{v}^{2}}={{u}^{2}}+2as$
where v= final velocity, s= distance travelled, u=initial velocity and a= acceleration
Complete step by step solution:
Here we consider v as the maximum velocity attained by the lorry to satisfy the condition and t is the minimum time taken by the lorry to reach from one station to another. Let t’ be the duration of acceleration and t” be the time of deceleration. From newton’s first equation of motion, we have velocity,
$v=0+at'$
From the question we can write that the total distance travelled by the lorry is:
$L=\dfrac{1}{2}at'+v(t-2t')+\dfrac{1}{2}at'$
$L=\dfrac{{{v}^{2}}}{a}+vt-\dfrac{2{{v}^{2}}}{a}=vt-\dfrac{{{v}^{2}}}{a}$
From this equation we can write an equation for time taken as:
Total time taken,
$t=\dfrac{L}{v}+\dfrac{v}{a}$
For finding the maximum velocity attained differentiating the above equation with respect to velocity. That is;
$\dfrac{dt}{dv}=-\dfrac{L}{{{v}^{2}}}+\dfrac{1}{a}$
We know that when time is minimum,
$\dfrac{dt}{dv}=0$
On applying this condition, we get maximum velocity as,
$-\dfrac{L}{{v_{\max}}^2}+\dfrac{1}{a} =0 $
$\therefore {{v}_{\max }}=\sqrt{La}$
Therefore, the answer is option C.
Notes: Here we have to remember that the conditions are given. So, we have to formulate such an equation that satisfies every condition. Then only we can derive it and equate it to zero to find the extreme condition.
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