Question

a train accelerates from 36 km per hour to 54 km per hour in 10 sec....now find the acceleration and the distance traveled by car​

Answer 1

$\boxed{\bf{ \orange{\bigstar Given \longrightarrow }}}$

• Initial velocity [u] = 36km/h
• Final velocity [v] = 54km/h
• Time [t] = 10 sec

$\boxed{\bf{ \green{\bigstar To\ Find \longrightarrow }}}$

• Acceleration [a]
• Distance traveled [s]

$\boxed{\bf{ \red{\bigstar Formulas \ used \longrightarrow }}}$

• $\sf{a=\dfrac{v-u}{t}}$
• $\sf{s= ut+\dfrac{1}{2} a t^2 }$

where,

a = acceleration

v= final velocity

u= initial velocity

t= time

s= distance traveled

$\boxed{\bf{ \blue{\bigstar SoLution \longrightarrow }}}$

Firstly change km/h to m/s

we know in 1 km= 1000m and 1 hr = 3600 sec

so from this,

Final velocity = $\sf{ 54km/h = 54\times \dfrac{5}{18} m/s = 15 m/s }$

Initial velocity = $\sf{ 36km/h = 36 \times \dfrac{5}{18} m/s = 10m/s }$

Firstly we will find the acceleration [a]

$\sf{a=\dfrac{v-u}{t}}$

put the value in the formula we get,

$\sf { a = \dfrac{15-10}{10} }$

$\boxed{\bf{ \bigstar a= 0.5m/s^2 \bigstar } }$

now we will find the distance traveled,

by using 2nd equation of motion,

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

put the given value in formula we get,

$\sf{s= 10\times 10 + \dfrac{1}{2} \times 0.5 \times 10^2}$

$\sf{ s= 100 + 0.5\times 50}$

$\sf{s= 100+ 25}$

$\boxed{\bf{\bigstar s=125m \bigstar }}$

therefore, acceleration = 0.5m/s²

and distance traveled = 125m

________________________________

Answer 2

Answer:

Initial velocity [u] = 36km/h

Final velocity [v] = 54km/h

Time [t] = 10 sec

★To Find⟶

Acceleration [a]

Distance traveled [s]

★Formulas used⟶

\sf{a=\dfrac{v-u}{t}}a=

t

v−u

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

2

1

at

2

where,

a = acceleration

v= final velocity

u= initial velocity

t= time

s= distance traveled

★SoLution⟶

Firstly change km/h to m/s

we know in 1 km= 1000m and 1 hr = 3600 sec

so from this,

Final velocity = \sf{ 54km/h = 54\times \dfrac{5}{18} m/s = 15 m/s }54km/h=54×

18

5

m/s=15m/s

Initial velocity = \sf{ 36km/h = 36 \times \dfrac{5}{18} m/s = 10m/s }36km/h=36×

18

5

m/s=10m/s

Firstly we will find the acceleration [a]

{a={v-u}{t}}a=

t

v−u

put the value in the formula we get,

\sf { a = \dfrac{15-10}{10} }a=

10

15−10

★a=0.5m/s 2★

now we will find the distance traveled,

by using 2nd equation of motion,

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

2

1

at

2

put the given value in formula we get,

{s= 10\times 10 +{1}{2} \times 0.5 \times 10^2}s=10×10+

2

1

×0.5×10

2

{ s= 100 + 0.5\times 50}s=100+0.5×50

{s= 100+ 25}s=100+25

★s=125m★

therefore, acceleration = 0.5m/s²

and distance traveled = 125m

________________________________