Question

a jet plane starts from rest at s=0 and it is subjected to the acceleration shown

Answer 1

a*s = 1/2 * 22.5 * 150 (Area of triangle formed in the graph

= 1687.5

v^2 - u^2 = 2as

u=0

V^2 = 2*1687.5

=3375

v = √3375

=15√15

=58 (approx)

Answer 2

Answer:

v=46.47m/s

Explanation:

(0,22.5),(150,0)

Equating line

(a−22.5)=22.5−0 /0−150  (5−0)

20(a−22.5)=−35

20a−450=−35

a=  −35+450 /20

v (dv/ds )=  −35+450 /20

v *∫0  vdv= 1 / 20  60 *∫0  (−35+450)ds

v*2/2=1/20(-35*2/2+450s)60*|0

v*2=1/10 (−5400+27000)

v=46.47m/s