Question
Question: A car accelerates from rest at a constant rate of \[\alpha \]for some time, after which it decelerat...
A car accelerates from rest at a constant rate of αfor some time, after which it decelerates at a constant rate of β to come to rest. If the total time elapsed is t second, then calculate
- the maximum velocity attained by the car, and
- the total displacement travelled by car in terms of α, β and t.
Solution
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 t1.
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.
α=t1vor t1=αv………………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.
β=t−t1vor t−t1=βv………………Equation (2)
By adding equation (1) and (2), we get,
t=v(α1+β1)
⇒v=(α+βαβ)t
Therefore, the maximum velocity attained by the car = v=(α+βαβ)t
b) Let ‘t1’ be the time for which it accelerates, ‘v’ be the velocity after this time.
Now, when the acceleration is α
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 } \\
t = {t_1} + {t_2} = [v/\alpha ] + [v/\beta ] = v[(\alpha + \beta )/\alpha \beta ] \\
\Rightarrow v = t[\alpha \beta /(\alpha + \beta )] \\
{v^2} = {0^2} + 2\alpha {S_1} \\
\Rightarrow {S_1} = \dfrac{{{v^2}}}{{(2\alpha )}} \\
{0^2} = {v^2} + 2\beta {S_2} \\
\Rightarrow {S_2} = \dfrac{{{v^2}}}{{(2\beta )}} \\
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] \\