Question

A car stard from rest and acquire a valocity 54 km /h in 2 second find the accelaration an distance travelled by assume motion of car ( pzz do this math for physics) pzz see the questions given 2 second ​

Answer 1

Concept :-

When a body starts from rest then its initial Velocity becomes zero and it is moving with final velocity only.

Given :-

Final Velocity of car = v = 54 Km/h = 15 ms-¹

Time = t = 2 s

We are required to find acceleration and distance transversed.

For Acceleration we will use first equation of motion.

v = u + at

15 = 2a

a = 7.5 ms-².

Now we have acceleration, then we can find distance traveled using second equation of motion as well as third equation of Motion. If time is not given then we use third equation of motion but in this case time is already given so we will use second equation of Motion.

S = ut + 1/2at²

S = 0 + 1/2 × 15/2 × 2 × 2

S = 15 m.

Again,

Finding the distance using third equation of Motion.

v² = u² + 2aS

(15)² = 0² + 2 × 15/2 × S

225 = 15S

S = 15 m

Hence,

Acceleration of body = a = 7.5 ms-²

Distance transversed = S = 15 m

Answer 2

$\huge \bigstar \rm { Correct \: Question}$

A car start from rest and acquire a velocity 54 km /h in 2 second, find the accelaration and distance travelled by assume motion of car .

$\huge \bigstar \rm { Solution }$

Given that,

➤ Initial velocity ( u ) = 0m/s

➤ Final velocity ( v ) = 54 km/h = 15m/s

⇢ Converting to its standard form -

$\tt {⇢ \cancel {54} \times \dfrac{5}{\cancel {18}} }$

$\tt {⇢3 \times 5}$

$\tt {⇢15m/s}$

➤ Time ( t ) = 2sec

Finding acceleration-

Formula -$\red \bigstar$ ${\boxed {\tt { a = \dfrac{v-u}{t} } }}$

Where,

➤ a ( acceleration ) = ?

➤ v ( final velocity ) = 15m/s

➤ u ( intial velocity ) = 0m/s

➤ t ( time ) = 2sec

$\tt { :\implies a = \dfrac{15-0}{2} }$

$\tt { :\implies a =\cancel { \dfrac{15}{2} } = 7.5m/{s}^{2} }$

$\rm \red { \therefore Acceleration \: is 7.5m/{s}^{2}}$

Finding distance travelled-

Formula - -$\red \bigstar$ ${\boxed {\tt { s = ut + \dfrac{1}{2}a{t}^{2} }}}$

Where,

➤ s ( distance travelled ) = ?

➤ u ( intial velocity ) = 0m/s

➤ t ( time ) = 2sec

➤ a ( acceleration ) = 7.5m/s^2

➤ t ( time ) = 2sec

$\tt { :\implies s = (0 \times 2 ) + \dfrac{1}{2} \times 7.5 \times {(2)}^{2} }$

$\tt { :\implies s = 0 + \dfrac{1}{\cancel {2}} \times 7.5 \times\cancel { 4 }}$

$\tt { :\implies s = 0 + \times 7.5 \times 2}$

$\tt { :\implies s = 15m}$

$\rm \red { \therefore Distance \: travelled \: is 15m}$