Dr. morgan is late for work. he enters the on-ramp to the merritt parkway at 15 m/s and
accelerates at a uniform 3.75 m/s2 up the 50 m ramp and for an additional 160 m on the
highway. he then sees a cop parked on the side of the road up ahead. he hits the brakes, causing him to slow at a rate of 2.50 m/s2, until he reaches the speed limit of 24.6 m/s (55 mph). determine the amount of time that passes between the moment dr. morgan
enters the on-ramp and until he reaches 55 mph (during his deceleration).

The amount of time is 14.44 seconds

Explanation:

* Lets explain how to solve the problem

- Dr.Morgan enters the on-ramp to the Merritt Parkway at 15 m/s

∴ His initial velocity u = 15 m/s

- He accelerates at a uniform 3.75 m/s² up the 50 m ramp and for

 an additional 160 m on the  highway

∴ The acceleration a = 3.75 m/s²

∴ The distance s = 50 + 160 = 210 m

- He hits the brakes, causing him to slow at a rate of 2.50 m/s²

 until he reaches the speed limit of 24.6 m/s

∴ The deceleration is 2.50 m/s²

∴ His final velocity v = 24 m/s

- We need to find the time of entire trip

* We have two stage in the problem:

# 1st stage:

- s = 210 m , u = 15 m/s , a = 3.75 m/s²

∵ s = ut + 1/2 at², where s is the distance, u is the initial velocity,

  a is the acceleration and t is the time

- Substitute the value of s , u and a in the equation above

∴ 210 = 15t + 1/2 (3.75)t²

∴ 210 = 15t + 1.875t²

- Subtract 210 from both sides

∴ 1.875t² + 15t - 210 = 0

- Solve the equation for t

∴ t = 7.31 seconds

# 2nd stage:

- d = 2.50 m/s² , v = 24.6 m/s

- The initial velocity in this stage is the final velocity of the

 1st stage, so lets calculate it

∵ v² = u² + 2as, where v is the final velocity, u is the initial velocity

   s is the distance and a is the acceleration

∴ v² = (15)² + 2(3.75)(210) = 1800

∴ v = √1800 = 42.43 m/s

∵ v in the 1st stage is u in the 2nd stage

∴ u = 42.43 m/s

- Now to fine the time of this stage we can use the rule

* Remember: d = -a

∴ seconds

∵ The time of the entire trip is the sum of the time of the 1st stage

  and the time of the 2nd stage

∵ t of 1st stage = 7.31 second

∵ t of 2nd stage = 7.13 seconds

∴ The total time = 7.31 + 7.14 = 14.44 seconds

∴ The amount of time is 14.44 seconds

the focus of lesson 1 is newton's first law of motion - sometimes referred to as the law of inertia. newton's first law of motion is often stated as. an object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force.

explanation:

gas and liquid