Mr. Turner is driving through downtown Houston at 35 MPH (15.56 m/s). This is above the posted speed limit and there are cars parked alongside the curb in the lane next to Mr. Turner. Of course, a large delivery truck ahead looks like it is parked and driverless, but the driver's door swings open into the lane ahead of Mr. Turner. Mr. Turner's reaction time is 3/4th's of a second and the stopping distance at his speed is 1.7 meters. The driver exiting the delivery truck is 13.3 meters ahead. Can Mr. Turner stop in time? What is the vehicle's acceleration when the brakes are applied? How far will the vehicle travel before Mr. Turner engages the brake?

(I need someone to explain this to me)

Mr. Turner will not be able to stop in time.

a = - 71.2 m/s²

s₁ = 11.67 m

Explanation:

Since, Mr. Turner is driving at 45.56 m/s and his reaction time for applying brakes is 3/4th of a second. So, the distance covered by car in this time will be:

s₁ = vt

where,

s₁ = distance covered by car before applying brakes = ?

v = speed of car = 15.56 m/s

t = reaction time = 3/4th of second = 0.75 s

Therefore,

s₁ = (15.56 m/s)(0.75 s)

s₁ = 11.67 m

and the distance required for the car to stop after applying brakes is:

s₂ = 1.7 m

So, the total distance traveled by car before stopping will be:

s = s₁ + s₂

s = 11.67 m 1.7 m

s =  13.37 m

since, the driver was 13.3 m ahead.

Therefore, Mr. Turner will not be able to stop in time.

To find the deceleration of Mr. Turner after applying brakes we use 3rd equation of motion:

2as₂ = Vf² - Vi²

where,

a = deceleration = ?

Vf = Final Velocity = 0 m/s

Vi = Initial velocity = 15.56 m/s

Therefore,

2a(1.7 m) = (0 m/s)² - (15.56 m/s)²

a = - (242.1136 m²/s²)/3.4 m

a = - 71.2 m/s²

a) cold, dry

explanation:

continental polar masses are cold and dry air masses because the cold air cannot hold much moisture. as they move into warmer air, they can trigger precipitation.