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Explanation & Answer
only Q 6 is left...i try and get back to you ...patel..please give some more time if you have
Question (1)
(a)
Object velocity (t ) =
ds (t )
= 6t 2 − 30t + 36 = 6( x − 2)( x − 3)
dt
(b)
Object acceleration a (t ) =
d (t )
= 12t − 30 = 6(2t − 5)
dt
(c)
From (a) and (b), we can see that velocities are zero at t = 2,3 and acceleration is zero
at t = 2.5
Hence at t = 2 , a 0 ; thus, there is a change in direction.
Hence at t = 3 , a 0 ; thus, there is a change in direction.
(d)
Sl.No.
1.
2.
3.
4.
Time Duration
0-2 units
2-2.5 units
2.5-3 units
>3
Speed
Slowing down
Speeding up
Slowing down
Speeding up
Question 2:
g ( x) =
2x 3 − x 2
x 2 (2 x − 1)
=
x 2 − 2 x − 3 ( x − 3)( x + 1)
The function is discontinuous at x = −1,3 . The function doesn't approach a particular
finite value at x = −1,3 ; thus, the limit does not exist. This is an infinite discontinuity.
Question 3:
(a) Given that
=
t2
ds
= sin
dt
20
t2
ds = sin dt
20
Location of the object at point is given by:
t2
s(t ) = sin dt
20
(b)
8
t2
s(t = 8) = sin dt = 3.999
20
0
(c)
My solution is ba...