Question

A car accelerates uniformly from 18 Kmh-1 to 36 Kmh-1 in 5s. calculate the acceleration and the acceleration and distance
covered by the car in that time ​

Answer 1

Given :

Initial velocity = 18km/hr

Final velocity = 36km/hr

Time interval = 5s

To Find :

Acceleration of car and distance covered.

Solution :

❖ Acceleration is defined as the rate of change of velocity$.$

• It is a vector quantity having both magnitude as well as direction.
• It can be positive, negative or zero.
• SI unit : m/s²
• Dimension formula : [L¹T‾²]

Formula : a = (v - u) / t

» u = 18 km/hr = 5 m/s

» v = 36 km/hr = 10 m/s

By substituting the given values;

➙ a = (10 - 5) / 5

➙ a = 5/5

➙ a = 1 m/s²

❒ Distance travelled by car :

Applying 3rd equation of kinematics;

➠ v² - u² = 2as

➠ 10² - 5² = 2(1)s

➠ 100 - 25 = 2s

➠ 2s = 75

➠ s = 75/2

➠ s = 37.5 m

Answer 2

$\huge{\underline{\underline {\mathtt{\purple{Hlw}\pink{Mate}}}}}$

$\Large\bold\blue{Given:}$

Initial velocity = 18km/hr

Final velocity = 36km/hr

Time interval = 5s

$\Large\bold\blue{To\:Find:}$

Acceleration of car and distance

$\Large\bold\blue{Solution:}$

Acceleration:

SI unit : m/s²

Dimension formula : [L¹T‾²]

Formula : a = (v - u) / t

u = 18 km/hr = 5 m/s

v = 36 km/hr = 10 m/s

Substituting the given values;

➙ a = (10 - 5) / 5

➙ a = 5/5

➙ a = 1 m/s²

Distance :

As per 3rd equation of kinematics;

v² - u² = 2as

10² - 5² = 2(1)s

100 - 25 = 2s

2s = 75

s = 75/2

s = 37.5 m

$\Large\fbox\red{Mark\:As\:Brainliest}$