Question

Velocity of car at $t = 0$ is $u$ , moves with a constant acceleration $\alpha $ and then decelerates at a constant rate $\beta $ for some time and gain its velocity, if the total time taken is $T$ , maximum velocity of car is given by

Answer 1

Hint: Here we have to use the equation of motion $v = u + at$ for first condition as well as for the second condition. After that find out the total time which can be calculated using the equation of motion. And then we have our velocity.

Complete step by step explanation:
For uniform acceleration, there are three equations of motion by Newton called the laws of constant acceleration or the laws of motion. These equations are used to derive the quantities like displacement(s), velocity (initial and final), time(t) and acceleration(a). Hence, these can only be used when acceleration is constant and motion is in a straight line. The three equation are
$v = u + at$
$s = ut + \dfrac{1}{2}a{t^2}$
${v^2} - {u^2} = 2as$
And here according to the question the car moves with a velocity $u$ at starting and has a constant acceleration $\alpha $ . And achieve a velocity $v$ in time ${t_1}$ .
So, by first equation of motion
$v = u + at$
Here $v$ is the final velocity and $u$ is the initial velocity.
Insert above values of acceleration and time
$v = u + \alpha {t_1}$
Solve for ${t_1}$ we get
${t_1} = \dfrac{{v - u}}{\alpha }$
Now after some time ${t_2}$ car moves with velocity $v$ and decelerates at constant rate $\beta $ and comes at its initial state of velocity$u$.
Again, from the first equation of motion,
$u = v + \left( { - \beta } \right){t_2}$ because it decelerates it have negative value
Solve for ${t_2}$ we getting
${t_2} = \dfrac{{v - u}}{\beta }$
Now total time $T$ is
$T = {t_1} + {t_2}$
Insert values of ${t_1}$ and ${t_2}$ , we get
$T = \dfrac{{v - u}}{\alpha } + \dfrac{{v - u}}{\beta }$
$ \Rightarrow \dfrac{{\left( {v - u} \right)\beta + \left( {v - u} \right)\alpha }}{{\alpha \beta }}$
By solving it,
$ \Rightarrow \dfrac{{v\left( {\alpha + \beta } \right) - u\left( {\alpha + \beta } \right)}}{{\alpha \beta }}$
Now cross multiplying, we get
$T\alpha \beta = v\left( {\alpha + \beta } \right) - u\left( {\alpha + \beta } \right)$
$ \Rightarrow T\alpha \beta + u\left( {\alpha + \beta } \right) = v\left( {\alpha + \beta } \right)$
Finally solve for $v$
$v = \dfrac{1}{{\alpha + \beta }}\left( {T\alpha \beta + u\left( {\alpha + \beta } \right)} \right)$
$v = \dfrac{{T\alpha \beta }}{{\alpha + \beta }} + u$

Note: Do not forget about velocities that are initial and what is final. Don’t confuse them with one another. Also remember deceleration means negative acceleration. When solving it always collect like terms.