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Basic knematics question i just cant get right

  • Thread starter devanlevin
  • Start date
  • #1
devanlevin
A car travelling ate velocity V starts to accelerate, in its 4th second of acceleration it passes 18m(ie during the 4th second, not after 4 seconds), in its 5th second it passes 22m , assuming the acceleration a is constant(stays the same) throughout what was the starting velocity V? how do i solve this



what i did was say,

a=delta v/delta t,

since a is constant and he passed 18 m in the 4th second(ie from t3-t4) and 22 in the 5th i conclude
v(t=3.5)= 18m/s
v(t=4.5)= 22m/s
=>delta v=4m/s
delta t = 1s
=>a=4m/s^2

the using V(t)=Vo+at
V(3.5)=Vo+4*3.5
=>18=Vo+14
=>Vo=4

is this a correct calculation or am i doing something wrong somewhere?

the whole question is in kinematics and there are no other forces afecting the car.

the answer in the book says Vo=0m/s and i cannot figure out how they got to that, it might be a mistake
 

Answers and Replies

  • #2
63
0
since a is constant and he passed 18 m in the 4th second(ie from t3-t4) and 22 in the 5th i conclude
v(t=3.5)= 18m/s
v(t=4.5)= 22m/s
=>delta v=4m/s
delta t = 1s
=>a=4m/s^2
This part seems to be wrong. Think about this...

I start accelerating from zero velocity and cover 10 meters in one second..does this mean my initial velocity v(t=0) is 10 m/s?
(No, because v(t=0)=0 since I started from rest)

Now try the question again..
 
  • #3
devanlevin
that i understand, but it does mean that your average velocity is 5, does it not, and thats why i said that the average velocity during the 4th second was 18, or in other words, since v is constant that the velocity exactly between the beginning and the end of the fourth second was 18.

do you see any other way to solve this and get a Vo of 0??
 
  • #4
tiny-tim
Science Advisor
Homework Helper
25,832
249
Welcome to PF!

A car travelling ate velocity V starts to accelerate, in its 4th second of acceleration it passes 18m(ie during the 4th second, not after 4 seconds), in its 5th second it passes 22m , assuming the acceleration a is constant(stays the same) throughout what was the starting velocity V? how do i solve this

since a is constant and he passed 18 m in the 4th second(ie from t3-t4) and 22 in the 5th i conclude
v(t=3.5)= 18m/s
v(t=4.5)= 22m/s
=>delta v=4m/s
delta t = 1s
=>a=4m/s^2

the using V(t)=Vo+at
V(3.5)=Vo+4*3.5
=>18=Vo+14
=>Vo=4

is this a correct calculation or am i doing something wrong somewhere?
Hi devanlevin! Welcome to PF! :smile:

Yes, that seems fine!

But either the question is badly worded, or you've read it wrong.

If it reads "in its 5th second of acceleration it passes 18m(ie during the 5th second, not after 4 seconds), in its 6th second it passes 22m", then Vo is 0! :smile:
 
  • #5
63
0
the using V(t)=Vo+at
V(3.5)=Vo+4*3.5
=>18=Vo+14
=>Vo=4
ok you used t= 3.5 s. Then its fine. I missed it last time.

yeah i don't see any problems now
 
  • #6
devanlevin
cant get it right, even using your equations still dont get Vo=0, think its a problem in the book??
 
  • #7
437
1
* it's a matter of concepts.


in its 4th second of acceleration it passes 18m(ie during the 4th second, not after 4 seconds)

velocity at this instant is therefore 18 m/s

in its 5th second it passes 22m

velocity at this instant is 22 m/s

in 1 sec, there is a change of velocity from 18 m/s to 22m/s. what is the acceleration?

using v = u + at

u = v - at

pick either velocity and its corresponding time and the acceleration you just calculated to get the initial velocity, u.
 
  • #8
63
0
cant get it right, even using your equations still dont get Vo=0, think its a problem in the book??
which problem are you talking about?
The answer to your problem is given as v0=0?
 
  • #9
62
0
Ok, my idea is this:

Set up two equations

d1 = V0*t1 + 1/2*a*(t1)^2
and
d2= V0*t2 + 1/2*a*(t2)^2


Now time one and time two are given (4s and 5s) and distance one and distance two are given as well(18m and 22m)

Therefore, we now have two equations in two variables which should not be a problem to solve. Please correct me if i'm wrong.
 
  • #10
63
0
Ok, my idea is this:

Set up two equations

d1 = V0*t1 + 1/2*a*(t1)^2
and
d2= V0*t2 + 1/2*a*(t2)^2


Now time one and time two are given (4s and 5s) and distance one and distance two are given as well(18m and 22m)

Therefore, we now have two equations in two variables which should not be a problem to solve. Please correct me if i'm wrong.
yes this would be the general method...
but i cant see anything wrong with his method..
taking averages is fine because velocity is linearly varying with time in this case
 
  • #11
62
0
I didn't say his method was wrong, i just made a suggestion :)
 

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