Moving car
A 1000kg car speeds up on the street at 72km/h. It suddenly brakes with a constant force of 5x103 N. How far the car will travel from the moment it brakes till it completely stops?
Answers
To work this out you will need to go through a number of steps of maths and logic. Take care and check as you go to avoid making a slip. First of all think if you have seen an equation linking force, mass and acceleration. One form of the equation that you might know is F = ma. Then you will need to re-arrange that equation to have just acceleration on one side.
19 March 2013
You will also have to think about units for the quantities. There is a force in Newtons given in the question so that is good. There is a mass given in kilograms and that is also good. There is a speed but not an acceleration and the speed is given in km/h but I would like to convert it into the standard unit for speed, metres per second or m/s.
19 March 2013
How do you convert 72 km/h into a speed in m/s? First think about the distance part of it and put 72 km into metres, easy if you know that kilo means thousand. So you now have 72000 metres per hour. The distance travelled in one second will of course be much less. The factor you need is the number of seconds in an hour.
19 March 2013
Also note that the question asks how far the car travels from when it starts to brake until it stops. The answer needs to be a distance. When you have worked out the speed of the car in m/s and the acceleration (negative acceleration or slowing down) you can work out the time taken for the breaking and then you can find the braking distance. Do let me know if you would like a private session to help you with the question.
22 March 2013
I put breaking instead of braking. You would not want me teaching you spelling would you?
22 March 2013
First of all we will check the data given to us that is in this case
20 August 2013
Time = t = ?
20 August 2013
What we have to find is
20 August 2013
Mass of car = M = 1000 kg ,
Force applied on brake for stoping car = F = 5x10^3 N ,
Initial Velocity at the time of application of brake (which is the final velocity of moving car) = Vi = 72 km/hr = 72 x (1000/3600) m/sec ,
Final Velocity = Vf = 0 m/sec
20 August 2013
There are three equation of motions , 1). Vf = Vi + a*t 2). S = Vit + (1/2)a*t^2 , 3). 2aS = Vf^2 - Vi^2where S is distance travelledNow we will use first equation because most of the variables in are known and it will give us what we need. There is one variable we need to find which is acceleration 'a'Using 2nd Law of Newton we will find acceleration which isF = M*aa = F/Ma = (5x10^3)/(1000) = 5 m/sec^2Putt this value of acceleration in 1st equation of motionVf = Vi + a*t0 = 20 + 5*tRearranging equation5*t = -20t = -20/5t = -4 secTime is never negative so ignore negative sign. So it will take 4 seconds by car to stop after the application of brakes.Negative sign come due to the decrease in acceleration and velocity by application of brake.
20 August 2013
Sorry question ask about distance and I find the time. For distance we will use 3rd equation of motion2aS=Vf^2-Vi^22(5)S = 0^2 - 20^2it is 20m/sec10*S = -400S = -40 magain answer of distance comes negative so ignore negative sign because distance never be negative. This is due to the deceleration. So car moves 40 meter after the application of brakes.
20 August 2013
Find acceleration using F=ma and then use a in one of the motion equations to find distance. Should try and then compare to number of answers given below.
10 January 2014
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