Rectilinear Motion help

  • #1
A car moving at 30 km/hr accelerates for 7.0 s. after which its velocity is 88 km/hr. How far does the car travel while its accelerating?

Given:
Initial Velocity : 30 km/hr
Final Velocity : 88 km/hr
Time : 7 seconds
Acceleration: ?
Distance: ?

correct me if i'm wrong

First I need to get the acceleration and convert my Vf and Vi to seconds. a = Vf - Vi / T


88km/hr - 30km/hr = 58 km/hr divide by 3600 so it will be in seconds so 0.016111111s / 7s = A

now that I have my acceleration (0.002301587286 km/s) I can either use this formula 2ad = Vf^2 - Vi^2 or D = Vi * T + 1/2 at^2

sooo...

(88 km/hr)^2 - (30 km/hr)^2 / 2(0.002301587286km/s) = Distance
so convert Vf^2 and Vi^2 to seconds again then divide by 2(0.002301587286km/s) to get distance

other solution :D

(30 km/hr) (7s) + (0.001150793643km/s) (7s)^2 = D

in this equation do I have to convert 30 km/hr to sec? If 2 of my equation are right both should have the same answer but both are different :(


Now my question is are all my equation/solution right? If not please correct it. :)
 

Answers and Replies

  • #2
Nathanael
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(30 km/hr) (7s) + (0.001150793643km/s) (7s)^2 = D

in this equation do I have to convert 30 km/hr to sec?
Yes, you do.
(or you could convert km/s to km/hr and the "7s" to "7/3600 hr" .... you just need to be consistent, however you choose to do it)

(88 km/hr)^2 - (30 km/hr)^2 / 2(0.002301587286km/s) = Distance
You also need to convert the "per hour" to "per second" in this equation (twice)



Doing this will give you consistent answers.
 
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  • #3
Nathanael
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Now, I may be wrong about this (someone correct me if I am,) but I think that this problem doesn't give enough information



First I need to get the acceleration and convert my Vf and Vi to seconds. a = Vf - Vi / T
The equation you wrote gives the AVERAGE velocity.
(It also happens to be the instantaneous velocity if and only if the acceleration is constant.)

But, the problem does not state the acceleration is constant... It just says "accelerates for 7 seconds"


So the distance covered will depend on HOW it accelerates throughout that 7 seconds.
(For example, if in the first second it is accelerated up to 87 km/hr, and in the last 6 seconds it only increases up to 88 km/hr, then the distance covered will be greater than your answer.)



So your answer is correct (once you make the units consistent and solve) ONLY IF the acceleration is constant (which is not in the problem statement)
 
  • #4
adjacent
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So your answer is correct (once you make the units consistent and solve) ONLY IF the acceleration is constant (which is not in the problem statement)
Yes, I agree.

An advice to the OP. Always convert all the units to SI units. Then it will make the problem easier.
Convert the km/h to m/s and use the equation of motion involving the distance, initial velocity, final velocity and time.
 
  • #7
I converted 7s to hours. Then I used both equation so I can check if two my answers are the same and I got 0.114722186 km :D
 
  • #8
adjacent
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I converted 7s to hours. Then I used both equation so I can check if two my answers are the same and I got 0.114722186 km :D
That is the right answer :smile:
 
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