Question

A car starts from rest and moves with uniform acceleration of $\alpha m{{s}^{-2}}$ along a straight line. It then retards uniformly at a rate $\beta m{{s}^{-2}}$ and stops. If $t$ is the time elapsed, then find the average speed of the car.

Answer 1

Average speed of the car is equal to the ratio of total distance covered by the car and the total elapsed time, taken to cover this total distance. Total distance covered by the car is equal to the sum of distances covered by the car while accelerating as well as decelerating. Distances covered by the car while accelerating and decelerating can easily be determined from the kinematic equations of motion.

Formula used:
$1){{v}_{av}}=\dfrac{{{d}_{tot}}}{{{t}_{tot}}}$
$2){{d}_{tot}}={{d}_{\alpha }}+{{d}_{\beta }}$
$3){{d}_{\alpha }}=u{{t}_{1}}+\dfrac{1}{2}\alpha {{({{t}_{1}})}^{2}}$
$4){{v}_{\alpha }}=u+\alpha {{t}_{1}}$
$5)v={{v}_{\alpha }}-\beta (t-{{t}_{1}})$
$6){{v}^{2}}-{{v}_{\alpha }}^{2}=2\beta {{d}_{\beta }}$

Complete step by step answer:
We are provided that a car starts from rest and moves with uniform acceleration of $\alpha m{{s}^{-2}}$ along a straight line. It then retards uniformly at a rate $\beta m{{s}^{-2}}$ and stops. If $t$ is the time elapsed, then, we are required to find the average speed of the car.
We know that average speed of the car is given by the ratio of total distance covered by the car and the total elapsed time. Mathematically, average speed of the car is given by
${{v}_{av}}=\dfrac{{{d}_{tot}}}{{{t}_{tot}}}=\dfrac{{{d}_{tot}}}{t}$
where
${{v}_{av}}$ is the average speed of the car
${{d}_{tot}}$ is the total distance covered by the car
${{t}_{tot}}=t$ is then total elapsed time, as provided
Let this be equation 1.
Now, if ${{d}_{\alpha }}$ and ${{d}_{\beta }}$ represent the distances covered by the car while accelerating and decelerating respectively, then, we can write the total distance covered by the car as
${{d}_{tot}}={{d}_{\alpha }}+{{d}_{\beta }}$
Let this be equation 2.
Now, from the second equation of motion, we have
${{d}_{\alpha }}=u{{t}_{1}}+\dfrac{1}{2}\alpha {{({{t}_{1}})}^{2}}=\dfrac{1}{2}\alpha {{({{t}_{1}})}^{2}}$
where
${{d}_{\alpha }}$ is the distance covered by the car while accelerating
$u=0$ is the initial velocity of the car, as the car starts from rest
${{t}_{1}}$ is the assumed time taken by the car to travel the distance ${{d}_{\alpha }}$
$\alpha$ is the acceleration of the car, as provided
Let this be equation 3.
Also, from the first equation of motion, we have
${{v}_{\alpha }}=u+\alpha {{t}_{1}}=\alpha {{t}_{1}}$
where
${{v}_{\alpha }}$ is the final velocity of the car just before decelerating
$u=0$ is the initial velocity of the car, as the car starts from rest
${{t}_{1}}$ is the assumed time taken by the car to travel the distance ${{d}_{\alpha }}$
$\alpha$ is the acceleration of the car, as provided
Let this be equation 4.
Similarly, from the first equation of motion, we have
$v={{v}_{\alpha }}-\beta (t-{{t}_{1}})\Rightarrow 0=\alpha {{t}_{1}}-\beta (t-{{t}_{1}})\Rightarrow \alpha {{t}_{1}}=\beta (t-{{t}_{1}})=\beta t-\beta {{t}_{1}}\Rightarrow {{t}_{1}}=\dfrac{\beta t}{\alpha +\beta }$
where
${{v}_{\alpha }}=\alpha {{t}_{1}}$ is the initial velocity of the car just after accelerating, from equation 4
$v=0$ is the final velocity of the car, since it stops
${{t}_{1}}$ is the assumed time taken by the car to travel the distance ${{d}_{\alpha }}$
$\beta$ is the deceleration of the car, as provided
$t-{{t}_{1}}$ is the time taken by the car to travel distance ${{d}_{\beta }}$
$\alpha$ is the acceleration of the car, as provided
Let this be equation 5.
Now, using the third equation of motion, we have
${{v}^{2}}-{{v}_{\alpha }}^{2}=2\beta {{d}_{\beta }}\Rightarrow 0-{{(\alpha {{t}_{1}})}^{2}}=2\beta {{d}_{\beta }}\Rightarrow {{d}_{\beta }}=\dfrac{-{{(\alpha {{t}_{1}})}^{2}}}{2\beta }$
where
$v=0$ is the final velocity of the car, as the car stops after deceleration
${{v}_{\alpha }}=\alpha {{t}_{1}}$ is the initial velocity of the car while decelerating
$\beta$ is the deceleration, as provided
${{d}_{\beta }}$ is the distance covered by the car while decelerating
Let this be equation 6.
Substituting equation 3, equation 5 and equation 6 in equation 2, we have
${{d}_{tot}}={{d}_{\alpha }}+{{d}_{\beta }}=\dfrac{1}{2}\alpha {{({{t}_{1}})}^{2}}-\dfrac{{{\alpha }^{2}}{{t}_{1}}^{2}}{2\beta }\Rightarrow {{d}_{tot}}=\dfrac{1}{2}\alpha {{({{t}_{1}})}^{2}}\left( 1-\dfrac{\alpha }{\beta } \right)=\left( \dfrac{\alpha }{2} \right){{\left( \dfrac{\beta t}{\alpha +\beta } \right)}^{2}}\left( 1-\dfrac{\alpha }{\beta } \right)$
Reducing the above expression further, we have
${{d}_{tot}}=\left( \dfrac{\alpha }{2} \right){{\left( \dfrac{\beta t}{\alpha +\beta } \right)}^{2}}\left( 1-\dfrac{\alpha }{\beta } \right)=\left( \dfrac{\alpha }{2} \right)\left( \dfrac{{{\beta }^{2}}{{t}^{2}}}{{{\left( \alpha +\beta \right)}^{2}}} \right)\left( \dfrac{\beta -\alpha }{\beta } \right)=\dfrac{\alpha \beta {{t}^{2}}}{2(\beta -\alpha )}$
Let this be equation 7.
Substituting equation 7 in equation 1, we have
${{v}_{av}}=\dfrac{{{d}_{tot}}}{t}=\dfrac{\dfrac{\alpha \beta {{t}^{2}}}{2(\beta -\alpha )}}{t}=\dfrac{\alpha \beta t}{2(\beta -\alpha )}$
Let this be equation 8.
Therefore, from equation 8, it is clear that the average speed of the car is equal to $\dfrac{\alpha \beta t}{2(\beta -\alpha )}$.

Note:
It is important to note that ${{v}_{\alpha }}$ used in the above solution is both the final velocity of the car once it has accelerated as well as the initial velocity of the car once it starts decelerating. This can be clearly understood from equations 4 and 5. Also, when we talk about deceleration, a negative sign in its value is important. This can be understood from equation 5.