Question

The velocity of a car increases from 30 km per hr to 45 km per hr in 5min .Calculate acceleration and distance travelled

Answer 1

Answer :-

Acceleration = 187.5km/hr^2

Distance = 3km

Explanation :-

Given :

• Initial velocity,u = 30km/hr

• Final velocity,v = 45km/hr

• Time,t = 5min => 5/60 = 0.08hr

[1min = 1/60hr]

To Find :

• Acceleration,a = ?

• Distance,s = ?

Solution :

According to the first equation of motion,

$\boxed{\sf{}v=u+at}$

Put their values and find for “a”

$\sf{}\implies 45=30+a\times 0.08$

$\sf{}\implies 45-30=a\times 0.08$

$\sf{}\implies 15=a\times 0.08$

$\sf{}\implies 15\div 0.08=a$

$\sf{}\implies 15\div \dfrac{8}{100}=a$

$\sf{}\implies 15\times\dfrac{100}{8}=a$

$\sf{}\implies 15\times\dfrac{50}{4}=a$

$\sf{}\implies 15\times\dfrac{25}{2}=a$

$\sf{}\therefore a=187.5km/hr^2$

Therefore,acceleration is equal to 187.5km/hr^2

According to the second equation of motion,

$\boxed{\sf{}s=u+\dfrac{1}{2}at^2}$

Put their values and find “s”

$\sf{}\implies s=30\times 0.08+\dfrac{1}{2}\times 187.5\times (0.08)^2$

$\sf{}\implies s=2.4+\dfrac{1}{2}\times 187.5\times0.0064$

$\sf{}\implies s=2.4+\dfrac{1}{1}\times 187.5\times0.0032$

$\sf{}\implies s=2.4+0.6$

$\sf{}\implies s=3km$

Therefore,distance travelled is equal to 3km

Answer 2

GiveN :

• Initial velocity (u) = 30 km/h
• Final velocity (v) = 45 km/h
• Time (t) = 5 mins = 0.084 h

To FinD :

• Acceleration
• Distance

SolutioN :

Use 1st equation of motion :

⇒v = u + at

⇒45 = 30 + a(0.084)

⇒a = (45 - 30)/0.084

⇒a = (15)/0.084

⇒a = 178.57 km/h²

_____________________

Use 3rd equation of motion :

⇒s = (v² - u²)/2a

⇒s = (45² - 30²)/(2*178.57)

⇒s = (2025 - 900)/357.14

⇒s = 1125/357.14

⇒s = 3.15 km