Question

The bus, at location A, is traveling at 30 m/s when it is shifted to neutral and allowed to

coast. Find the speed of the bus at location B. **Hint: Let location B be o potential

energy. h is the height above this point.

Answer 1

Complete Question

The complete question is shown on the first uploaded image

Answer:

The velocity of the bus at B is
`v_B= 31.6\ m/s`

Step-by-step explanation:

Let's take position B as base point .

From the diagram height between point B and A ia mathematically evaluated as

`h_A = 60 -55`

`h_A = 5 \ m`

From the question we are told that

The velocity at location A is
`v_A = 30 m/s`

According the law of conservation of energy

`PE_A + KE_A = KE_B`

Where
`PE_A` is the potential energy at A which is mathematically represented as

`PE_A = mgh_A`

`KE_A` is the kinetic energy energy at A which is mathematically represented as

`KE_A = (1)/(2) mv^2_A`

`KE_B` is the kinetic energy at A which is mathematically represented as

`KE_B = (1)/(2) mv_B^2`

Where
`v_B` is the velocity at location B

So

`mgh_A + (1)/(2) mv_A^2 = (1)/(2) mv_B^2`

Making
`v_B` the subject of the formula

`v_B= \sqrt{(gh_A + (1)/(2) v_A^2 )/(0.5) }`

Substituting values

`v_B= \sqrt{((9.8 * 5) + (1)/(2) 30^2 )/(0.5) }`

`v_B= 31.6\ m/s`