Deceleration. My answer is different from the books and it's driving me crazy

  • Thread starter laxboi33
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  • #1
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Homework Statement


The driver of a pickup truck going 100 km/h applies the brakes, giving the truck a uniform deceleration of 6.50 m/s^2 while it travels 20.0 m.

a) what is the speed of the truck in kilometers per hour at the end of this distance?

b) How much time has elapsed?


Homework Equations





The Attempt at a Solution




The first thing I did was convert 100 km/h into m/s and got 277.7 m/s.

After converting I tried using V^2= V(initial)^2 x 2(a)x, where a = acceleration and x = horizontal distance.

I got

V^2 = (277.7 m/s)^2 + 2 (-6.50 m/s^2) (20.0-0)
V^2 = 77117.29 + (-260)
V^2 = 76857.29
V= 277.23 m/s which equates to 99.8 km/hr which makes no sense.

I then tried x = x(initial) + 1/2 at^2 to try and acquire time first.

I ended up getting 83 seconds. Which is way off. So I checked the answers in the back of the book and it read:

for part a) 81.4 km/h
b) .794 s

can someone please tell me what I'm doing wrong? I'm obviously not asking for the answer since I've got the book.
 

Answers and Replies

  • #2
nasu
Gold Member
3,785
438
You have an error in your conversion in m/s. It's more like 27 m/s.
Just divide 100 km/h by 3.6.
 
  • #3
turin
Homework Helper
2,323
3
The first thing I did was convert 100 km/h into m/s and got 277.7 m/s.
I think that you're off by a decimal place.
 
  • #4
66
0
x = x(initial) + v(initial)(t) + 1/2at^2....correct?
 
  • #5

Homework Statement


The driver of a pickup truck going 100 km/h applies the brakes, giving the truck a uniform deceleration of 6.50 m/s^2 while it travels 20.0 m.

a) what is the speed of the truck in kilometers per hour at the end of this distance?

b) How much time has elapsed?


Homework Equations





The Attempt at a Solution




The first thing I did was convert 100 km/h into m/s and got 277.7 m/s.

After converting I tried using V^2= V(initial)^2 x 2(a)x, where a = acceleration and x = horizontal distance.

I got

V^2 = (277.7 m/s)^2 + 2 (-6.50 m/s^2) (20.0-0)
V^2 = 77117.29 + (-260)
V^2 = 76857.29
V= 277.23 m/s which equates to 99.8 km/hr which makes no sense.

I then tried x = x(initial) + 1/2 at^2 to try and acquire time first.

I ended up getting 83 seconds. Which is way off. So I checked the answers in the back of the book and it read:

for part a) 81.4 km/h
b) .794 s

can someone please tell me what I'm doing wrong? I'm obviously not asking for the answer since I've got the book.






Sorry your very first step of converting kmph to mps is wrong. A small mistake in decimal point .

1km= 1000m
1 hr= 3600s

100kmph= (100*1000)/3600 = 1000/36
=27.777m/s
 
  • #6
Sorry your very first step of converting kmph to mps is wrong. A small mistake in decimal point.

1km= 1000m
1 hr= 3600s

100kmph= (100*1000)/3600 = 1000/36
=27.777m/s

http://arasaraja.blogspot.com/" [Broken]
 
Last edited by a moderator:
  • #7
76
0
You need to use the equation d = Vot + at^2/2 (with d = 20), you know Vo and a so if you solve it you'll receive 2 solutions for t, only one is valid
 

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