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Appropriate Question :-

If a body travels with uniform acceleration 5 m/s for 5 sec and with uniform acceleration 10 m/s² for next 10 sec then its average acceleration in m/s² is?

Given :-

Initial velocity of the body = 5 m/s
Initial time of the body = 5 sec
Final time of the body = 10 sec
Final acceleration = 10 m/s²

To Find :-

The Average acceleration = ?

Solution :-

At first assume that ;

$\quad { \bullet { \sf { Average \: Acceleration \: = A_{avg} } } }$

$\quad { \bullet { \sf { Initial \: velocity \: = u } } }$

$\quad { \bullet { \sf { Initial \: time \: in \: which \: it \: attains \: a \: velocity \: "u" = t_{1} } } }$

$\quad { \bullet { \sf { Final \: time \: attain \: acceleration \: of \: 10 m/s² \: = \: t_{2} } } }$

$\quad { \bullet { \sf { Acceleration \: by \: Final \: time \: t_{2} \: = \: A } } }$

$\quad { \bullet { \sf { Final \: velocity \: = v } } }$

By Given ;

u = 5 m/s
${ \sf { t_{1} = 5 } }$
${ \sf { t_{2} = 10 } }$
${ \sf { A = 10 m/s² } }$
${ \sf { v = ? } }$

It is clear that the final acceleration is the change velocity within the given time , numerically ;

${ \sf { \bf { A = \dfrac{v - u}{t_{2}} } } }$

Putting the values we have ;

$: \longmapsto { \sf { \bf { 10 = \dfrac{v - 5}{10} } } }$

$: \longmapsto { \sf { \bf { v - 5 = 10 × 10 } } }$

$: \longmapsto { \sf { \bf { v - 5 = 100 } } }$

$: \longmapsto { \sf { \bf { v = 100 + 5 } } }$

$: \longmapsto { \sf { \bf { v = 105 m/s } } }$

Now as Average velocity is the rate of change velocity upon rate of change of time . So ;

${ \sf { \bf { A_{avg} = \dfrac{v - u}{t_{2} - t_{1}} } } }$

Now , Putting all the values we have ;

$: \longmapsto { \sf { \bf { A_{avg} = \dfrac{105 - 5}{10 - 5} } } }$

$: \longmapsto { \sf { \bf { A_{avg} = \dfrac{100}{5} } } }$

$: \longmapsto { \sf { \bf { A_{avg} = 20 m/s² } } }$

Henceforth , Our Required Answer is 20 m/s² .

lohitjinaga

Answer:

The initial velocity travelled is 625m

The accleration is α=10m/s

The time is t=5s

Apply the equation of motion

v=u+at

The velocity is v=0+10.5=50m/s.

The distance travelled in the first 5s is

s=ut+

=0.5+

.10.5

=125m

Thedistancetravelledinthefollowing10sis

=ut

=50.10=500m

Thetotaldistancetravelledis

d=s+s

=125+500=625m

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