The relation between time t and distance x is t = ax 2 + bx 3 , where a and b are
constants. Find the instantaneous acceleration.
Answers
Answer:
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Explanation:
ANSWER
t=ax
2
+bx
Differentiate w.r.t time
1=2ax
dt
dx
+b
dt
dx
dt
dx
=
2ax+b
1
V=
2ax+b
1
...(1)
a=
dt
dV
=
(2ax+b)
2
−(2a
dt
dx
)
a=
(2ax+b)
2
−2av
....(2)
Substiting
2ax+b
1
=V in (2)
we will get, a=−2av
3
Answer:
answer : -2av³
explanation : The relation between time t and distance x is given by, t = ax² + bx, where a and b are constants.
let's differentiate t with respect to x,
i.e., dt/dx = d(ax² + bx)/dx
or, dt/dx = 2ax + b .....(1)
we know, velocity is the rate of change of displacement with respect to time.
i.e., v = dt/dx
from equation (1),
dt/dx = 1/{dx/dt} = 1/v = 2ax + b
or, v = 1/(2ax + b) ......(2)
now differentiating v with with respect to time, t
dv/dt = d{1/(2ax + b)}/dt
= -1/(2ax + b)² × d(2ax + b)/dt
= -1/(2ax + b)² × [2a × dx/dt ]
= -1/(2ax + b)² × 2a v
from equation (2),
dv/dt = -v² × 2av = -2av³
we know, acceleration/retardation is the rate of change of velocity with respect to time.
i.e., A = dv/dt
so, dv/dt = A = -2av³
[here negative sign shows retardation.]
hence, answer is -2av³
Explanation: