Question

A bus of mass 5000kg starts from rest and rolls down a hill if it travel a distance of 200m calculate (i) acceleration (ii) force acting on the bus​

Answer 1

Correct Question :-

☛ A bus of mass 5000kg starts from rest and rolls down a hill if it travel a distance of 200m in 10s.calculate

(i) acceleration

(ii) force acting on the bus

Given :-

• Mass of bus = 5000kg

• Initial velocity = 0m/s

• Distance covered by bus =200m

• Time taken = 10s

To Find:-

• (i) Acceleration

• (ii) Force acting on bus.

Solution :-

(i)

By using 1st equation of motion

➦ v = u+at

➭ v = 0+ a × 10

➭ v = 10a m/s

∴The final velocity of bus is 10a m/s

and Now,

Using 3rd equation of motion

➦ v² = u² +2as

➭( 10a)² = 0² + 2 ×a ×200

➭ 100a² = 400a

➭ 100a = 400

➭ a = 400/100

➭ a = 4m/s²

∴ The acceleration of bus is 4m/s².

(ii)

We know that,

➦ Force = mass× acceleration

➭ Force = 5000×4

➭ Force = 20000N

∴ The force acting on the bus is 20000N.

Answer 2

Answer:

acceleration (a) = 4m/s²

force (f) = 20 kN

Explanation:

It starts from rest,

so initial velocity (u) = 0

final velocity (v) = ?

Distance (s) = 200m

time (t) = 10s

(i) acceleration

let the acceleration (a) = a m/s²

we know that ,

from first equation of motion ,

⇒ v = u +at

⇒ = 0 + a×10

⇒ = 10a m/s

we know that ,

From third equation of motion

⇒ v² - u² = 2as

⇒ (10a)² - 0 = 2×a×200

⇒ 100a² = 400a

⇒ 100a = 400

⇒ a = 400/100 

⇒ a = 4 m/s²

(ii) force acting on the bus

mass = 5000kg

Force(F) = mass(m) ×acceleration (a)

= 5000×4

= 20,000 N

= 20 kN

here ,

kN = kilo Newton