Solutions (^109)
We discard the second root of the equation since
it has no physical meaning.
The velocity v of the helicopter at the instant
when the engine is switched off can be found from
the relation
c (^) c
at t. a [V(
)2 ,
a— rg •-•'- —
a (^) a j
= c 2 ±2act 2 — c = 80 m/s.
1.4. During a time t 1 , the point mass moving with
an acceleration a will cover a distance s = at;/2
and will have a velocity v = ati. Let us choose the
x-axis as shown in Fig. 121. Here point 0 marks
s=a 42 /2
Fig. 121
the beginning of motion, and A is the point at
which the body is at the moment t 1. Taking into
account the sign reversal of the acceleration and
applying the formula for the path length in uni-
formly varying motion, we determine the time t 2
iii which the body will return from point A to
point 0:
at 2
- rati2 t—
2 2 '
whence t 2 = ti (1 +
The time elapsed from the beginning of motion
to the moment of return to the initial position can
be determined from the formulat = ti+ ts=-- ti ( 2 + V-2)•
1.5. We shall consider the relative motion of the
bodies from the viewpoint of the first body. Then
at the initial moment, the first body is at rest (it
can be at rest at the subsequent instants as well),1 "( u =at,
0 A