Question

Calculate the first and second velocities of the car with two washers attached to the pulley, using the formulas

v1 = 0.25 m / t1, and

v2 = 0.25 m / (t2 – t1)

where t1 and t2 are the average times the car took to reach the 0.25 and the 0.50 meter marks. Record these velocities, to two decimal places, in Table E.
What is the first velocity of the car with two washers at the 0.25 meter mark?

m/s

What is the second velocity of the car with two washers at the 0.50 meter mark?

m/s

Answer 1

Calculation of velocities.

Explanation:

• Speed changed after a specific amount of time. This is represented by speed = d1/t1.

• Replacing values, we get 65 mi/h = (130 mi)/t1

• Meanwhile, the value of t1 is 2.0 h.

• Hence during lower speed the time is represented as such t2 = T - t1 = 3.33 h — 2.0 h = 1.33 h.

• This also gives us the difference between both speeds such as speed = d2/t2 hence d2 = 73 mi.

• Further more, D = d1 + d2 = 130 mi + 73 mi = 203 mi.

• Normal speed = d/t = (203 mi)/(3.33 h) = 61 mi/h.

• Since the speed is not normal i.e 65 mi/h + 55 mi/h.

Answer 2

Answer:

Explanation:

Time passed before the speed change

speed = d1/t1 ;

65 mi/h = (130 mi)/t1

t1 = 2.0 h.

the time at the lower speed is

t2 = T — t1 = 3.33 h — 2.0 h = 1.33 h.

We discover the separation went at the lower speed from

speed = d2/t2;

55 mi/h = d2/ (1.33 h), which gives d2 = 73 mi.

The complete separation =

D = d1 + d2 = 130 mi + 73 mi = 203 mi.

normal speed = d/t = (203 mi)/(3.33 h) = 61 mi/h.

Note that the normal speed isn’t!

(65 mi/h + 55 mi/h). The two rates were not kept up for

break even with times