Initially you are driving at 45 ft/sec on a straight road

  • Thread starter Tonia
  • Start date
In summary, during the acceleration process from 45 ft/sec to 76 ft/sec over a distance of 300 ft, the car has an average speed of 60.5 ft/sec and an acceleration of 6.3 ft/sec^2. The time required for the 300 ft travel is 4.9 seconds. To reach a speed of 55.8 ft/sec, it takes the car 1.7 seconds. The initial velocity accounts for 69.3 ft of the 300 ft displacement, while the acceleration accounts for the remaining 230.7 ft. The time required for traveling the first 150 ft is approximately 2.1 seconds.
  • #1
Tonia
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Homework Statement


Initially you are driving at 45 ft/sec on a straight road. You accelerate the car at a steady rate and increase the speed to 76ft/sec. During this acceleration process you travel a distance of 300 ft.

Homework Equations


a) Calculate the time required for the 300 ft travel
b) determine the acceleration of the car during the 300 ft displacement
c) find the time when the speed of the car will be 55.8 ft/sec.
d) determine how much of the 300 ft displacement is due to the initial velocity and how much is due to the acceleration
e) find the time required for traveling the first 150 ft.

The Attempt at a Solution


a) time = distance divided by speed, I am not sure how to use this equation since I need to know the speed to find the time. Speed = distance divided by time.
b) I need help
c) I need help acceleration = m = a = change in velocity divided by change in time = (76ft/sec - 45 ft/sec)/time = ??
d) I need help
e) I need help
 
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  • #2
This problem builds on itself. You can't get part b unless you get part a.
a) time = distance divided by AVERAGE speed. What's the average of 45 and 76?
 
  • #3
(76ft/sec + 45 ft/sec.)/2 = 121/2 = 60.5 ft/ sec.
300 ft / 60.5 ft/ sec. = 4.9 seconds??
acceleration = m = a = change in velocity divided by change in time = (76ft/sec - 45 ft/sec)/4.9 sec. = 31ft/sec/4.9 sec. = 6.3 ft/sec^2

c) find the time when the speed of the car will be 55.8 ft/sec. ??
d) determine how much of the 300 ft displacement is due to the initial velocity and how much is due to the acceleration ??
e) find the time required for traveling the first 150 ft. ??
 
  • #4
You are traveling 45 ft/s. You are gaining 6.3 ft/s each second. How many seconds until you are doing 55.8 ft/s?
Think of it as a money problem. You have 45 dollars. You make $6.30 an hour. You need $55.8. How many hours must you work?
 
  • #5
55.8 ft/sec - 45 ft/sec = 10.8 ft/sec
10.8ft/sec/6.30ft/sec^2 = 1.714 sec. = 1.7 seconds

d) determine how much of the 300 ft displacement is due to the initial velocity and how much is due to the acceleration
e) find the time required for traveling the first 150 ft.
 
  • #6
d) How far would you have traveled at your initial velocity during the time you computed in part (a) if you did not accelerate? The rest belongs to acceleration.
e) x = vit + (1/2) at2
You could use the quadratic formula to solve for t. There may be easier ways.
 

1. How far will you travel in 10 seconds?

In 10 seconds, you will travel 450 feet (45 ft/sec x 10 sec = 450 feet).

2. What is your average speed in miles per hour?

Your average speed would be approximately 30.68 miles per hour (45 ft/sec x 0.681818 = 30.68 mph).

3. How long will it take to travel 1 mile?

It will take approximately 26.07 seconds to travel 1 mile (1 mile x 5280 feet/mile = 5280 feet / 45 ft/sec = 117.3333 sec = 117.3333 sec x 0.222222 = 26.07 sec).

4. How does your speed change if you accelerate at a rate of 5 ft/sec^2?

If you accelerate at a rate of 5 ft/sec^2, your speed will increase by 5 ft/sec every second. So after 1 second, your speed would be 50 ft/sec, after 2 seconds it would be 55 ft/sec, and so on.

5. What is your displacement after 5 seconds?

Your displacement after 5 seconds would be 225 feet (45 ft/sec x 5 sec = 225 feet). This means you would have traveled 225 feet in a straight line from your initial position.

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