Question

A particle starts with initial speed u and retardation a to come to rest in time T. The time taken to cover first half of the total path travelled is

plz solve this question with the help of graph not calculation

plzzzz​

Answer 1

EXPLANATION.

• GIVEN

particle starts with initial speed u

retardation a to come to rest in time T

To find time taken to cover first half of the

total path travelled is.

According to the question,

From Newton first equation of kinematics,

Initial speed = a

acceleration = ( -a)

time = T

final speed = 0

V = U + AT

0 = u - at

u = at .... (1)

To find distance,

From newton third equation of kinematics

${v}^{2} - {u}^{2} = 2as$

0 - u^2 = - 2as

s = u^2 / 2a ...... (2)

If it's distance be halved

$\frac{s}{2} = ut + \frac{1}{2}a {t}^{2}$

acceleration = - a

$\frac{s}{2} = ut - \frac{1}{2}at {}^{2}$

put the value in equation we get,

$\frac{u {}^{2} }{4a} = ut - \frac{1}{2}a {t}^{2}$

$\frac{a {}^{2} {T}^{2} }{4a} = aTt - \frac{1}{2}a {t}^{2}$

$2 {t}^{2} - 4Tt + {T}^{2}$

$t \: = \frac{ 4T \pm\sqrt{16 {T}^{2} - 8 {T}^{2} } }{4}$

$t \: = T(1 - \frac{1}{ \sqrt{2} } )$

Answer 2

Answer:

EXPLANATION.

GIVEN

particle starts with initial speed u

retardation a to come to rest in time T

To find time taken to cover first half of the

total path travelled is.

According to the question,

From Newton first equation of kinematics,

Initial speed = a

acceleration = ( -a)

time = T

final speed = 0

V = U + AT

0 = u - at

u = at .... (1)

To find distance,

From newton third equation of kinematics

{v}^{2} - {u}^{2} = 2asv

2

−u

2

=2as

0 - u^2 = - 2as

s = u^2 / 2a ...... (2)

If it's distance be halved

\frac{s}{2} = ut + \frac{1}{2}a {t}^{2}

2

s

=ut+

2

1

at

2

acceleration = - a

\frac{s}{2} = ut - \frac{1}{2}at {}^{2}

2

s

=ut−

2

1

at

2

put the value in equation we get,

\frac{u {}^{2} }{4a} = ut - \frac{1}{2}a {t}^{2}

4a

u

2

=ut−

2

1

at

2

\frac{a {}^{2} {T}^{2} }{4a} = aTt - \frac{1}{2}a {t}^{2}

4a

a

2

T

2

=aTt−

2

1

at

2

2 {t}^{2} - 4Tt + {T}^{2}2t

2

−4Tt+T

2

t \: = \frac{ 4T \pm\sqrt{16 {T}^{2} - 8 {T}^{2} } }{4}t=

4

4T±

16T

2

−8T

2

t \: = T(1 - \frac{1}{ \sqrt{2} } )t=T(1−

2

1

)