Question

a car travelling at velocity of 10 mis due north speeds up uniformuly to a velocity of 25 mls in 35 sec calculate the acceleration during this period​

Answer 1

Correct Question:

A car travelling at velocity of 10 m/s due north speeds up uniformly to a velocity of 25 m/s in 35 sec. Calculate the acceleration during this period.

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Given:

Initial velocity, u= 10 m/s due north

Final velocity, v= 25 m/s due north

Time taken, t= 35 s

To Find:

Acceleration during given time period ​

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

s is displacement

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Let the acceleration of car for given time period be 'a'

On applying first equation of motion on given car, we get

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

$\longrightarrow\rm{25=10+a(35)}$

$\longrightarrow\rm{25-10=a(35)}$

$\longrightarrow\rm{15=35a}$

$\longrightarrow\rm{35a=15}$

$\longrightarrow\rm{a=\dfrac{\cancel{15}}{\cancel{35}}}$

$\longrightarrow\rm{a=\dfrac{3}{7}}$

$\longrightarrow\rm\green{a=0.42\:m/s^{2}\:due\:north}$

Hence, the acceleration of car for given time period is 0.42m/s² due north.

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Additional Information-

• According to second equation of motion for constant acceleration,

$\orange{\boxed{\bf{s=ut+\dfrac{1}{2}at^{2}}}}$

• According to third equation of motion for constant acceleration,

$\pink{\boxed{\bf{v^{2}-u^{2}=2as}}}$

where,

v is final velocity

u is initial velocity

a is acceleration

s is displacement

Answer 2

Question :-

A car travelling at velocity of 10 m/s due north speeds up uniformly to a velocity of 25 m/s in 35 seconds . Calculate the acceleration during this period ?

Answer :-

Given :-

A car is travelling at velocity of 10 m/s due north speeds up uniformly to a velocity of 25m/s in 35 seconds

Required to find :-

• Acceleration during this period ?

Concept used :-

• The definition of acceleration

Formula used :-

$\large{\leadsto{\boxed{\rm{ Acceleration = \dfrac{ \Delta v }{ \Delta t }}}}}$

Solution :-

Given :-

A car is travelling at velocity of 10 m/s due north speeds up uniformly to a velocity of 25m/s in 35 seconds

So, from the above statement we can conclude that,

Initial velocity ( u ) = 10 m/s

Final velocity ( v ) = 25 m/s

Time = 35 seconds

He asked us to find the acceleration of the car

So,

What is acceleration ?

Acceleration is the change in velocity per unit time .

So,

This is represented as ,

$\rm{ Acceleration = \dfrac{ Change \; in \; velocity }{Change \; in \; time }}$

$\implies{\mathsf{Acceleration = \dfrac{ \Delta v }{ \Delta t }}}$

Here,

This is " ∆ " known as Delta

This " ∆ " represents change .

So,

Here,

Change in the velocity is the difference between the final velocity and initial velocity .

∆v = v - u

So,

.

Using this above concept let's solve this question

Hence,

$\tt{ Acceleration = \dfrac{ v - u }{t}}$

Substitute the respective values

$\Rightarrow{\tt{ Acceleration = \dfrac{ 25 \; m/s - 10 \; m/s }{ 35 }}}$

$\Rightarrow{\tt{ Acceleration = \dfrac{ 15 \; m/s }{ 35 \; s }}}$

$\implies{\tt{ Acceleration = \dfrac{ 3 \; m/s}{ 7 \; s}}}$

$\implies{\tt{ Acceleration = 0.4285 - - - - }}$

$\implies{\tt{ Acceleration = 0.42 \; m/s^2 ( approximately )}}$

$\large{\leadsto{\boxed{\rm{\therefore{ Acceleration \; during \; this \; period = 0.42 \; m/s^2 }}}}}$

Points to remember :-

1. The S.I unit of acceleration is m/s .

2. Actually, the first equation of motion was been derived using this above formula .

So, you can solve this using the 1st equation of motion or by using the above formula .

3. The three equations of motion are ,

$\large{\tt{ 1.\; v = u + at }}$

$\large{\tt{2. \; s = ut + \dfrac{1}{2}at^2 }}$

$\large{\tt{3. \; v^2 - u^2 = 2as }}$

Here,

v = final velocity

u = initial velocity

s = displacement

t = time

a = acceleration