Question

1. Ramesh writes his cocotter with a speed of 72
km pli hour and applies breaks and come to
stop in 10 seconds find acceleration distance
Covered
by him before he stop​

Answer 1

Initial speed of the scooter is 72 km/hr and after applying brakes final velocity becomes 0 km/hr. The scooter comes to rest in 10 seconds.

We have to find the acceleration and distance covered by Ramesh.

To convert km/hr into m/s. Multiply with 5/18.

Initial speed = 72*5/18 = 20 m/s and final velocity = 0 m/s.

Using the First Equation Of Motion:

v = u + at

Substitute the known values,

→ 0 = 20 + a(10)

→ -20 = 10a

→ -2 = a

(Negative sign shows retardation)

Therefore, the acceleration of the scooter is 2 m/s².

Now, for distance

Using the Third Equation Of Motion:

v² - u² = 2as

Substitute the values,

→ (0)² - (20)² = 2(-2)(s)

→ 0 - 400 = -4s

→ 100 = s

Therefore, the distance covered by the scooter is 100 m.

Answer 2

QUESTION :

Ramesh rides his scooter with a speed of 72 km per hour and applies breaks and come to stop in 10 seconds find acceleration distance Covered by him before he stops

GIVEN :

• initial velocity

= 72 km / hr = 72 × 5 /18 = 20 m / s

( in order to convert it into m / s multiply it by 5 / 18 )

• final velocity = 0
• time taken = 10 sec

FORMULA :

• v = u + a t

here ,

1. v = final velocity
2. u = initial velocity
3. a = acceleration
4. t = time taken

• s = ut + at ² / 2

here,

1. s = distance

SOLUTION :

v = u + a t

0 = 20 + a 10

a = - 20 / 10

a = - 2 m / s ²

( Note : here negative acceleration represents decceleration )

s = u t + 1× a × t ²/ 2

s = 20×10 - 2 × 100 / 2

s = 200 - 100

s = 100 m

ANSWER :

• acceleration = - 2 m / s ²
• distance = 100 m