Question

A car accelerates uniformly from 18 kmh-¹ to 36 kmh-¹ in 5s. calculate (i). the acceleration and(ii). the distance covered by the car in that time .

Answer 1

u= 18×5/18 = 5m/s
v= 36×5/18= 10m/s
t= 5s
acceleration =v-u/t= 10-5/5= 1m/s^2
v^2-u^2 = 2as
100-25= 2×1 × distance
75/2 = distance covered.
34.5 m is the distance covered.

Answer 2

Answer:

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