An object moves with constant acceleration 4.20 m/s2 and over a time interval re
ID: 2992159 • Letter: A
Question
An object moves with constant acceleration 4.20 m/s2 and over a time interval reaches a final velocity of 12.6 m/s. (a) If its initial velocity is 6.3 m/s, what is its displacement during the time interval? m (b) What is the distance it travels during this interval? m (c) If its initial velocity is -6.3 m/s, what is its displacement during the time interval? m (d) What is the total distance it travels during the interval in part (c)?Explanation / Answer
a) v^2=u^2+2*a*D =>(12.6)^2=(6.3)^2+2*(4.8)*S => S = 12.403125m b) here distance is same as displacement, therefore distance = 12.403125m c)v^2=u^22*a*D =>(12.6)^2=(-6.3)^2+2*(4.8)*S => S = 12.403125m same as (a) since displacement is calculated(it depends only on initial and final position i.e. direction and not the path used). d) v1=0, v2=12.6 with v=v1; v^2=u^2+2*a*D =>(0)^2=(-6.3)^2+2*(4.8)*S => S = -4.134375m D1=|S|= 4.134375m with v=v2,u=v1; v^2=u^2+2*a*D =>(12.6)^2=(0)^2+2*(4.8)*S => S = 16.5375m D2=|S|=16.5375m total distance = D1+D2 = 4.134375m+16.5375m = 20.671875m