a drag racer, starting from rest speeds up for 402m with an acceleration of 17. a parachute then opens?

Question

slowing the accelatation to -6.10 how fast the racer moving 3.50*10^2 after parachute opens? thankx

Answer 1

Just draw the V-T diagram. From (0,0) V rises in a straight line somewhere in +x-y quadrant. Then all of sudden parachute opens which bends the graph in tent shaped figure approaching the x-axis,

let t1 be the time in sec during which the racer accelerated and reached final velocity V

u=0 m/s, s=402 m, f= 17 m/s^2
then s= 0 +0.5 f t1^2 = 402
>>> t1 = 6.877 sec
V = 0 + ft1 >>>
V1 = 116.909 m/s was achieved after 402 distance

when parachute opened, let us assume that "this switch-over of acceleration into deceleration" was instantaneous and this can be treated as an abrupt boundary. thus let us treat this point as new t' =0

such that V(t'=0) = V1=116.909 m/s as the racer continues to move with deceleration of (- a = 6.10) (negative) then its velocity will continue to decrease with time or space covered.
at S (t') = 3.5*100 meter (unit missing in question)
V(t') will be given by

V^2 (t') = V1^2 - 2* a S(t')
V^2 (t') = 13667.71 - 4270 = 9397.71

V (t') = 96.94 m/s

Answer 2

That a hard one to answer