# You are traveling on an interstate highway at the posted speed limit of 70 mph when you see that the traffic in front of you has stopped due to an accident up ahead. You step on your brakes to slow down as quickly as possible. Assume that you to slow down to 30 mph in about 5 seconds. A) With this same average acceleration, how much longer would it take you to stop?B) What total distance would you travel from when you first apply the brakes until the car stops?

A.The time taken for the car to stop is 8.75 s

B.The distance travelled when the brakes were applied till the car stops is 136.89 m

A. Determination of the time taken for the car to stop.

• We'llbegin bycalculatingthedecelerationof thecar

Initial velocity (u) = 70 mph = 0.447 × 70 = 31.29 m/s

Final velocity (v) = 30 mph = 0.447 × 30 = 13.41 m/s

Time (t) = 5 s

### a = –3.576 m/s²

• Finally,we shall determine the time taken for the car to stop.

Initial velocity (u) = 31.29 m/s

Final velocity (v) = 0 m/s

Deceleration (a) = –3.576 m/s²

### t = 8.75 s

Thus, the time taken for the car to stop is 8.75 s

B.Determination of the total distance travelled when the brakes were applied.

Initial velocity (u) = 31.29 m/s

Final velocity (v) = 0 m/s

Deceleration (a) = –3.576 m/s²

### s = 136.89 m

Therefore, the total distance travelled by the car when the brakes were applied is 136.89 m

136.89 m

Explanation:

Given

Initial velocity

velocity after 5 s is

Therefore acceleration during these 5 s

therefore time required to stop

v=u+at

here v=final velocity =0 m/s

initial velocity =31.29 m/s

(b)total distance traveled before stoppage

s=136.89 m

## Related Questions

Listed following are locations and times atwhich different phases of the Moon are visible fromEarth’s....? Listed following are locations and times atwhich different phases of the Moon are visible from Earth’sNorthern Hemisphere. Match these to the appropriate moon phase.

The three given moon phases are:

Waxing Crescent Moon

Waning Crescent Moon

Full Moon

The things we need to match to the above three topicsare:

*visible near western horizon an hour after sunset
*rises about the same time the sun sets
*visible near eastern horizon just before sun rises
*occurs about 3 days before new moon (i know this is waningcrescent)
*visible due south at midnite
*occurs 14 days after new moon (i know this is full moon)

Explanation:

*visible near western horizon an hour aftersunset    WAXING CRESCENT

*rises about the same time the sun sets    FULLMOON

*visible near eastern horizon just before sun rises  WANING CRESCENT

*occurs about 3 days before new moon WANING CRESCENT

*visible due south at midnite     FULL MOON

*occurs 14 days after new moon FULL MOON

Explain what happent to the pressure exerted by an object when the area over which it is exerted:a) increase
b) decrease​

The pressure decreases if the area increases. If the area decreases then the pressure increases.

The 160-lblb crate is supported by cables ABAB, ACAC, and ADAD. Determine the tension in these wire

Explanation:

Using the diagram (see attachment) we extract the following position vectors:

Next step is to find unit vectors as follows:

Using the diagram we find the corresponding vectors Forces:

Equation of Equilibrium:

Comparing i , j and k components as follows:

Solving Above equation simultaneously we get:

1 cm = 100 m
1 mm = 100 cm
100 mm = 1 cm
1 m = 100 cm

The last one

1m = 100 cm

Explanation:

If you do not trust me look it up

A tuning fork vibrates at 15,660 oscillations every minute. What is the period (in seconds) of one back and forth vibration of the tuning fork?

We are given:

The tuning fork vibrates at 15660 oscillations per minute

Period of one back-and forth movement:

the given data can be rewritten as:

1 minute / 15660 oscillations

60 seconds / 15660 oscillations          (1 minute  = 60 seconds)

dividing the values

0.0038 seconds / Oscillation

Therefore, one back and forth vibration takes 0.0038 seconds

At what distance from a long straight wire carrying acurrentof 5.0A is the magnitude of the magnetic field due to the
wireequal to the strength of the Earth's magnetic field of about
5.0 x10^-5 T?

The distance is 2 cm

Solution:

According to the question:

Magnetic field of Earth, B_{E} =

Current, I = 5.0 A

We know that the formula of magnetic field is given by:

where

d = distance from current carrying wire

Now,

d = 0.02 m 2 cm