Question

Madhulika drive scooter acquires a velocity of 36 kilometre per hour in 10 seconds just after the start. she takes 20 seconds to stop. calculate the acceleration in two cases

Answer 1

Given:

Case-1: When scooter starts

Initial velocity of scooter,u= 0 km/h

(Since, scooter starts from rest)

Final velocity of scooter,v= 36 km/h

Time taken by scooter,t= 10 s

Case-2: When scooter stops

Initial velocity of scooter,u'= 36 km/h

Final velocity of scooter,v'= 0 km/h

(Since, scooter stops finally)

Time taken by scooter,t'= 20 s

To Find:

Acceleration in both the cases

Solution:

We know that,

• According to first equation of motion for constant acceleration,

$\purple{\boxed{\bf{v=u+at}}}$

where,

v is final velocity

u is initial velocity

a is acceleration

t is time taken

$\rule{190}{1}$

It is given that,

$\rm{u=v'=0\ km/h=0\times\dfrac{5}{18}\ m/s=0\ m/s}$

$\rightarrow\rm{v=u'=36\ km/h=36\times\dfrac{5}{18}\ m/s=10\ m/s}$

$\rule{190}{1}$

Case-1

Let the acceleration of scooter be a

So, on applying first equation of motion on scooter, we get

$\longrightarrow\rm{v=u+at}$

$\longrightarrow\rm{10=0+a(10)}$

$\longrightarrow\rm{10=10a}$

$\longrightarrow\rm{10a=10}$

$\longrightarrow\rm{a=\dfrac{10}{10}}$

$\longrightarrow\rm\green{a=1\ m/s^{2}}$

$\rule{190}{1}$

Case-2

Let the acceleration of scooter be a'

So, on applying first equation of motion on scooter, we get

$\longrightarrow\rm{v'=u'+a't'}$

$\longrightarrow\rm{0=10+a'(20)}$

$\longrightarrow\rm{-10=20a'}$

$\longrightarrow\rm{20a'=-10}$

$\longrightarrow\rm{a'=\dfrac{-10}{20}}$

$\longrightarrow\rm\green{a'=-0.5\ m/s^{2}}$

Answer 2

Question:

$Madhulika\:drive\:scooter\:acquires\:a\:velocity\\ of\:36\:kilometre\:per\:hour\:in\:10\:seconds\:just\:after\\ the\:start.\:she takes\:20\:seconds\:to\:stop. </p><p>\\calculate\:the\:acceleration\:in\:two\: cases.$

Given:

$Let\:the\:case\:one\:be\:C_{1}\\ And\:case\:two\:be\:C_{2}$

Case 1:

☞ $Initial\:velocity = 0 km\:hr^{-1} = 0 m\:s{-1}$

☞ $Final\:velocity = 36 km\:hr^{-1} = 10 m\:s{-1}$

☞ $Time\:Taken = 10 sec$ =

Case 2:

☞ $Initial\:velocity = 36 km\:hr^{-1} = 10 m\:s{-1}$

☞ $Final\:velocity = 0 km\:hr^{-1} = 10 m\:s{-1}$

☞ $Time\:Taken = 20 sec$

We Know:

$\blue{\boxed{v = u + at}}$

In the equation;

• u = Initial velocity
• v = final velocity
• a = accelaration
• t = Time taken

Reference:

❥In the First case the acceleration will be taken as positive.

❥In the second case the acceleration will be negative.

Solution:

For first case :

${\boxed{v = u + at}}$

$\blue{\boxed{10 m\:s^{-1} = 0 m\:s^{-1} + a × 10 sec}}$

$\blue{\boxed{10 = 10a}}$

$\blue{\boxed{a = \dfrac{10}{10} m\:s^{-2}}}$

$Hence,\:the\:acceleration\:is 1 m\:s^{-2}$

______________________________________

For second case :

${\boxed{v = u + at}}$

$\blue{\boxed{0 m\:s^{-1} = 36 m\:s^{-1} + a × 20 sec}}$

$\blue{\boxed{0 = 10 + 20a}}$

$\blue{\boxed{-10= 20a}}$

$\blue{\boxed{a = \dfrac{-10}{20} m\:s^{-2}}}$

$Hence,\:the\:acceleration\:is\:-0.5 m\:s^{-2}$

______________________________________