Question

A scooter accelerates from rest for time\[{t_1}\] at constant rate \[{\alpha _1}\;\]and then retards at constant rate \[{\alpha _{2\;}}\] for time\[{t_2}\] and comes to rest. The correct value of \[{t_1}/{t_2}\;\] will be
a. \[({\alpha _1}\; + {\text{ }}{\alpha _2})/{\alpha _2}\]
b. \[{\alpha _2}/{\alpha _1}\]
c. \[({\alpha _1}\; + {\text{ }}{\alpha _2})/{\alpha _1}\]
d. \[{\alpha _1}/{\alpha _2}\]

Answer 1

Hint: It is given that scooter accelerates at a constant rate means that that it does not change with time. In other words, velocity is increasing uniformly. This is the case of uniform acceleration. Uniform acceleration is given by change of velocity per unit time. There are two cases of the question. First is when the scooter is accelerating and other is retardation that is negative acceleration.

Complete answer:
Case\[1\]:
The scooter is starting from rest. Thus, initial velocity is zero. Thus, final velocity per unit time gives the value of acceleration.

Substituting all the data in the form of equation, we write:
\[{\alpha _1} = \frac{v}{{{t_1}}}\] ……….(1)
Where,
\[{\alpha _1}\] is the acceleration and v is the final velocity reached in time \[{t_{1.}}\]

Case\[2\]:
The scooter retards for time \[{t_2}\] at a rate \[{\alpha _{2.}}\]
Thus, retardation is given by negative of difference of final and initial velocity. Or it is correct to say that it is initial velocity minus the final velocity. The final velocity of case first becomes the initial velocity of the case second. Thus, initial velocity is v. Also, the scooter finally comes to rest so final is zero.

Thus in form of mathematical equation
\[{\alpha _2} = \frac{v}{{{t_2}}}\] ……..(2)

Dividing the two equations to get ratio $\frac{{{t_1}}}{{{t_2}}}$
$\frac{{{t_1}}}{{{t_2}}} = \frac{{{\alpha _2}}}{{{\alpha _1}}}$

Thus, the correct option is b.

Note: A graph can be plotted depicting the above situation. As scooter starts from rest, its velocity increases uniformly. Thus, a straight line starting from zero that has increasing slope. Then the scooter retards so a straight line with decreasing slope till zero because the scooter again comes to rest position.