Question

2. Calculate the distance covered by a car that starts from rest and attains velocity of 10m/s after an acceleration of 6m/s2.

Answer 1

Answer:

25/3 m

Explanation:

Given,Initial velocity(u)=0

v=10m/s,a=6/ms2

we know by 3rd equation of motion,

$v^{2} -u^{2} =2as\\2(6)s=10^{2} -0^{2} \\12s=100\\s=100/12\\s=25/3m$

Distance covered = 25/3 m

Answer 2

Given:-

→ Initial velocity of the car = 0

[as it starts from rest]

→ Final velocity of the car = 10m/s

→ Acceleration = 6m/s²

To find:-

→ Distance covered by the car

Solution:-

Let's calculate the distance covered by the car by using the 3rd equation of motion :-

=> v² - u² = 2as

Where:-

• v is final velocity of the body.

• u is initial velocity of the body.

• a is acceleration of the body.

• s is distance covered by the body.

Substituting values in the above equation, we get:-

=> (10)² - 0 = 2×6×s

=> 100 = 12s

=> s = 100/12

=> s = 8.33m [approximately]

Thus, distance covered by the car is 8.33m .

Some Extra Information:-

The 3 equations of motion for a body moving with uniform acceleration are:-

• v = u + at

• s = ut + 1/2at²

• v² - u² = 2as