Question

A car accelerates from rest at a constant rate of \[\alpha \]for some time, after which it decelerates at a constant rate of \[\beta \] to come to rest. If the total time elapsed is t second, then calculate
1) the maximum velocity attained by the car, and
2) the total displacement travelled by car in terms of \[\alpha \], \[\beta \] and t.

Answer 1

Hint: When the speed of an object increases along the direction in which it travels, aka it gets faster if we look at the direction in which it is moving, we say it accelerates. And the rate at which this "speeding-up" happens is called acceleration. Deceleration is the reverse, as the body slows down in its direction of motion.

Complete step-by-step answer:
So, assume a body speeds up first, then decelerates. That means it accelerates for a while first and is then made to slow down. That the highest possible speed should be?
The speed it achieves by the end of the acceleration, of course! And after that it starts slowing down and any speed value the body has will be lower than the speed at the end if it is.

a) The car accelerates from rest for a time \[{t_1}\].

Its velocity-time graph is therefore a straight-line OA sloping in the upward direction. The car's maximum velocity is considered 'v.'

The graph OA 's slope gives the car an acceleration.

\[\alpha = \dfrac{v}{{{t_1}}}\;or{\text{ }}{t_1} = \dfrac{v}{\alpha }\]………………Equation (1)
Upon achieving the maximum velocity, the car starts to decelerate. Therefore, the velocity-time graph is a straight-line AB sloping in the downward direction. The slope of the AB graph gives the car a detard.

\[\beta = \dfrac{v}{{t - {t_1}}}\;or{\text{ }}t - {t_1} = \dfrac{v}{\beta }\]………………Equation (2)

By adding equation (1) and (2), we get,

\[t = v\left( {\dfrac{1}{\alpha } + \dfrac{1}{\beta }} \right)\]
\[ \Rightarrow v = \left( {\dfrac{{\alpha \beta }}{{\alpha + \beta }}} \right)t\]
Therefore, the maximum velocity attained by the car = \[v = \left( {\dfrac{{\alpha \beta }}{{\alpha + \beta }}} \right)t\]

b) Let ‘\[{t_1}\]’ be the time for which it accelerates, ‘v’ be the velocity after this time.

Now, when the acceleration is \[\alpha \]
\[\begin{gathered}
  v = 0 + \alpha {t_1} \\
   \Rightarrow {t_{_1}} = \dfrac{v}{\alpha } \\
\end{gathered} \]

Then it decelerates at rate \[\beta \] and stops at \[{t_{_2}}\].

Hence,

\[
   \Rightarrow 0 = v - \beta {t_{_2}} \\
   \Rightarrow v = \beta {t_{_2}} \\
   \Rightarrow {t_{_2}} = \dfrac{v}{\beta } \\
\]

Now,

\[
  t = {t_1} + {t_2} = [v/\alpha ] + [v/\beta ] = v[(\alpha + \beta )/\alpha \beta ] \\
   \Rightarrow v = t[\alpha \beta /(\alpha + \beta )] \\
\]

Distance travelled in time \[{t_1}\] is found as,

\[

  {v^2} = {0^2} + 2\alpha {S_1} \\
   \Rightarrow {S_1} = \dfrac{{{v^2}}}{{(2\alpha )}} \\
\]

Distance travelled in time \[{t_2}\] is found as,

\[
  {0^2} = {v^2} + 2\beta {S_2} \\
   \Rightarrow {S_2} = \dfrac{{{v^2}}}{{(2\beta )}} \\
\]

The total displacement travelled by car is,
\[
  S = {S_1} + {S_2} \\
   \Rightarrow S = \dfrac{{{v^2}}}{{(2\alpha )}} + \dfrac{{{v^2}}}{{(2\beta )}} \\
   \Rightarrow S = \left[ {\dfrac{{{v^2}}}{2}} \right] + \left[ {\dfrac{1}{\alpha } + \dfrac{1}{\beta }} \right] \\
   \Rightarrow S = \left[ {\dfrac{{{v^2}}}{2}} \right] + \left[ {\dfrac{{\alpha + \beta }}{{\alpha \beta }}} \right] \\
   \Rightarrow S = \dfrac{1}{2}{\left[ {t\left[ {\dfrac{{\alpha \beta }}{{\alpha + \beta }}} \right]} \right]^2}\left[ {\dfrac{{\alpha + \beta }}{{\alpha \beta }}} \right] \\
   \Rightarrow S = \dfrac{1}{2}t\left[ {\dfrac{{\alpha \beta }}{{\alpha + \beta }}} \right] \\
\]

Hence the total displacement travelled by car in terms of \[\alpha \], \[\beta \] and t is \[\dfrac{1}{2}t\left[ {\dfrac{{\alpha \beta }}{{\alpha + \beta }}} \right]\]

Note: The significance of an object's velocity can be defined as the rate of change in the location of the object in relation to a frame of reference and time. It may sound complicated but velocity in a particular direction is essentially speeding. It is a quantity of vectors which means that we need both magnitude (speed) and direction to define velocity. Its SI unit is meter per second when there is a change of magnitude or direction in a body's velocity the body is said to be accelerating.