Question

please answer urgently with explanation​

Answer 1

$\bigstar\huge\boxed{\mathtt{\red{Solution:}}}$

Let us consider that the car travels from point A to point B with const acceleration α and from point B to point C with constant acceleration β

PART-I

As per the given question ,

As the car accelerates from rest the initial velocity of the car will be zero

Acceleration of the car = α

Time taken = t

As the acceleration of the car is constant we can use the first kinematic equation in order to solve this question

we know that,

$\rightarrow\mathtt{\green{v=u+at}}$

here,

• v= final velocity
• u= initial velocity
• a= acceleration
• t = time taken

$\rightarrow\mathtt{v=0+\alpha t_1}$

$\rightarrow\mathtt{v=\alpha t_1}$

$\rightarrow\mathtt{ t_1=\dfrac{v}{\alpha} }$

-------------------------------

PART-II

The final velocity of the car in PART-I will be the initial velocity of the car in PART-II

Initial velocity of the car = v

As the car comes to rest  the final velocity of the car will be zero

Time taken by the car to come to rest = t 2

Acceleration of the car = - β

Note : the negative sign is due to deceleration

As the acceleration of the car is constant we can use the first kinematic equation in order to solve this question

we know that,

$\rightarrow\mathtt{0=v-\beta t_2}$

$\rightarrow\mathtt{v=\beta t_2}$

$\rightarrow\mathtt{ t_2=\dfrac{v}{\beta} }$

----------------------------

The total time lapsed is t

we know that ,

$\rightarrow\mathtt{ t= t_1+t_2 }$

$\rightarrow\mathtt{ t= \dfrac{v}{\alpha} +\dfrac{v}{\beta} }$

$\rightarrow\mathtt{ t= \dfrac{v\alpha+v\beta}{\alpha\beta} }$

$\rightarrow\mathtt{ t= v\dfrac{(\alpha+\beta)}{\alpha\beta} }$

$\rightarrow\mathtt{ \pink{v=\dfrac{\alpha\beta}{\alpha+\beta}t }}$