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

  • Thread starter Tonia
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1. 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.

2. 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.

3. 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
 

tony873004

Science Advisor
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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?
 
96
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(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. ??
 

tony873004

Science Advisor
Gold Member
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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?
 
96
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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.
 

tony873004

Science Advisor
Gold Member
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141
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.
 

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