A particle moving along a straight line with a constant acceleration of -4m/s*2 passes through a point A with a velocity of +8m/s at some movement find the distance travelled by the particle in 5sec after that movement?

1 Answer
Jane
Jan 15, 2018

#26# meters

Explanation:

Let us make a graph of speed#(v)# vs time#(t)#

so,our equation becomes #v=|8-4t|# (using #v=u-at#) (mod taken because speed can never be negative)

so,putting values we get,
at,# t=0,v=8#
# t=1,v=4#
# t=2.v=0#
# t=3,v=4#
# t=4,v=8#
# t=5,v=12#

now,the following graph is made using the values obtained.enter image source here

hence, distance = area under this speed time curve,
i.e #1/2*8*2+1/2*12*3# or,#26# meters

#P.S # YOU JUST NEED TO KNOW THE INITIAL SPEED#(u)#,FINAL SPEED#(v)#,THE GIVEN TIME#(t2)# AND THE TIME IN WHICH SPEED BECOMES ZERO#(t1)# FROM THE GIVEN EQUATION,BECAUSE THESE VALUES ARE ONLY REQUIRED FOR CALCULATING AREA UNDER THE CURVE IN SUCH CASE,NO NEED OF FINDING ALL THE VALUES

so,your formula becomes, distance covered in# t2# seconds =
#1/2*u*t1 + 1/2*v*(t2-t1)#

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