Differentiate 3 Alpha^2 Beta with respect to X
Answers
Answer:
The relation between time and the distance is given by
t=αx
2
+βx
Differentiate the above equation with respect to time,
1=2xα
dt
dx
+β
dt
dx
. . . . . . .(1)
Differentiate with respect to time,
0=2α(x
dt
2
dx
2
+
dt
dx
dt
dx
)+β
dt
2
dx
2
. . . . .(2)
Velocity, v=
dt
dx
. . . . . . .(3)
Acceleration, a=
dt
2
d
2
x
. . . . . . . .(4)
Substitute equation (3) and (4) in equation (1), we get
0=2α(xa+v
2
)+βa
βa=−2α(xa+v
2
)=−2αxa−2αv
2
βa+2αxa=−2αv
2
a=−
β+2αx
2αv
2
. . . . . . . . .(5)
Substitute equation (3) and (4) in equation (1), we get
1=2xαv+βv
2xαv=1−βv
2αx=
v
1−βv
. . . . . . . . .(6)
Substitute equation (6) in equation (5), we get
a=−
β+
v
1−βv
2αv
2
=−
βv+1−βv
2αv
2
×v
a=−2αv
3
The retardation is 2αv
3
.
The correct option is A.
Answer:
Differentiate 3 Alpha^2 Beta with respect to X - 20083220.