View Full Version : A car weighs 1000Kg and the driver applies the brakes with a force of 400N.
Prostaff
Apr 27, 2012, 05:21 AM
Its physics. I got an answer but I'm not sure if I'm right
ebaines
Apr 27, 2012, 06:02 AM
Its physics. I got an answer but I'm not sure if I'm right
Your question got cut off - what are you trying to calculate, the car's rate of deceleration? Also - please tell us what answer you got and how you arrived at it, and we can tell you whether it's correct or not.
Prostaff
Apr 27, 2012, 09:27 PM
Your question got cut off - what are you trying to calculate, the car's rate of deceleration? Also - please tell us what answer you got and how you arrived at it, and we can tell you whether it's correct or not.
Yeah, I'm trying to find the car's rate of deceleration. I got
Time=2.5
Velocity due to gravity=10m/s
So deceleration=-(0-10/2.5)=4m/s^2
ebaines
Apr 30, 2012, 05:46 AM
To find the acceleration of the car use F=ma: F is the force applied by the brakes in the direction opposite of travel (hence is a negative number), and m is the mass of the car. So from F=ma you can rearrange to get: a = F/m = -400N/1000Kg.
Why diid you include 'g' (acceleration due to gravity? For a car on a level road (which I assume this is) gravity acts perpendicular to the direction of motion and hence has no effect on the car's acceleration.
Prostaff
May 1, 2012, 02:18 AM
To find the acceleeation of the car use F=ma: F is the force applied by the brakes in the direction opposite of travel (hence is a negative number), and m is the mass of the car. So from F=ma you can rearrarnge to get: a = F/m = -400N/1000Kg.
Why diid you include 'g' (acceleration due to gravity? For a car on a level road (which I assume this is) gravity acts perpendicular to the direction of motion and hence has no effect on the car's acceleration.
Thanks for the reply. I understand it now, it was on my test and I obviously screwed it up. Thanks though