A 1200 kg car moving at 25 m/s has the brakes applied with a deceleration of 8.0 m/s². How far does the car travel before it stops?
Answer (2)
v=vnaut+at 0(want it to stop)=25+-8t(negative accel cuz its deceling) -25/-8=t t=3.125 seconds or with sig figs 3.1 seconds x=xnaut+vnaut time+1/2 a time^2 x(where it ends up)=0(we start distance from when it started to brake)+25 3.1+1/2*-8 3.1^2=39.06 meters or 39.1 meters x=0+25 3.1-.5 8 3.1^2 is the same equation as above with nothing written in. Are you in physics ap or reg? either way friend me lol(just joined website tonight)
The car travels approximately 39.1 meters before coming to a stop, calculated using the kinematic equation for motion. The initial speed is 25 m/s, with a deceleration of 8 m/s². Using the equation v 2 = v 0 2 + 2 a x , we find that the distance traveled, x , is about 39.1 meters.
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