A particle is moving along a straight line such that its position depends on time as x=1-at+b^2t?,
where a = 2 m/s, b= 1 m/s then distance covered by the particle during first 3 seconds from starting
of the motion
Answers
Explanation:
Part a):
v
(
t
)
=
3
2
t
2
+
5
t
+
1.219
(
m/s
)
Part b):
Δ
x
=
16.438
m
Explanation:
Note: I apologize in advance for the long solution; I am not sure how much one who reads this understands and have attempted to give some explanation for each step taken. Hope this helps!
Let's begin by writing out what we are given:
a
(
t
)
=
3
t
+
5
(
m/s
2
)
v
(
0
)
=
−
4
(
ft/s
)
Before we begin to solve the problem, we need to convert acceleration and velocity into some common unit. In this solution, we will be using SI (
m
,
m/s
,
m/s
2
)
Since
1
m
is equal to
3.281
ft
, we can use some simple dimensional analysis to convert
v
(
0
)
from
ft/s
into
m/s
:
v
(
0
)
=
−
4
ft
s
⋅
1
m
3.281
ft
=
1.219
m/s
Remember that
a
=
d
v
d
t
, so:
d
v
=
a
d
t
Integrating both sides, we have:
∫
d
v
=
∫
a
d
t
So,
v
=
∫
a
d
t
Applying the above formula, we can find
v
(
t
)
in units of
m/s
:
v
(
t
)
=
∫
(
3
t
+
5
)
d
t
Here, we must remember how to integrate a polynomial:
∫
(
k
x
n
)
d
x
=
k
x
n
+
1
n
+
1
+
C
Since
a
(
t
)
=
3
t
+
5
is a polynomial, we can apply this integration method, so:
v
(
t
)
=
∫
(
3
t
+
5
)
d
t
=
3
2
t
2
+
5
t
+
C
Since we are given that
v
(
0
)
=
1.219
m/s
, we can solve for
C
:
v
(
0
)
=
3
2
(
0
)
2
+
5
(
0
)
+
C
=
1.219
Therefore,
C
=
1.219
and the answer to part a) is
v
(
t
)
=
3
2
t
2
+
5
t
+
1.219
m/s
To answer part b), we use the
v
(
t
)
equation found in part a) to find the point(s) at which the particle stops and turns around by setting
v
(
t
)
=
0
and solving for the roots of the quadratic
However, we can take a shortcut and notice that in the quadratic, both the
t
coefficient and constant are positive, which would thus result in only negative roots. Therefore, we do not need to worry about solving for
t
's at which
v
(
t
)
=
0
since we are only being asked to find the total distance traveled from
t
∈
[
0
,
2
]
Here, we need to remember that
v
=
d
x
d
t
, solving for
x
by a similar process as we did for
v
earlier:
v
=
d
x
d
t
d
x
=
v
d
t
∫
d
x
=
∫
v
d
t
x
=
∫
v
d
t
Since we have that
v
(
t
)
=
3
2
t
2
+
5
t
+
1.219
, we an substitute and solve the definite integral:
Δ
x
=
∫
2
0
(
3
2
t
2
+
5
t
+
1.219
)
d
t
Δ
x
=
(
3
6
t
3
+
5
2
t
2
+
1.219
t
)
]
2
0
Δ
x
=
(
1
2
t
3
+
5
2
t
2
+
1.219
t
)
]
2
0
Using the Fundamental Theorem of Calculus, we have that:
Δ
x
=
(
1
2
(
2
)
3
+
5
2
(
2
)
2
+
1.219
(
2
)
)
−
(
1
2
(
0
)
3
+
5
2
(
0
)
2
+
1.219
(
0
)
)
Plugging this into a calculator we have:
Δ
x
=
16.438
m
Answer: Distance covered by the particle during first 3 seconds from starting of the motion is 5 m
Explanation:
x = 1 - 2t + t²
v = dx/dt
v = -2 + 2t
v = 2(t-1)
Therefore,
for t < 1 , v < 0, particle moves in direction of negative axis
for t > 1 , v > 0 , particle moves in direction of positive axis.
The particle starts from x = 1 at t = 0. For first second it travels towards negative x axis and reaches x = 0 at t = 1 sec. The particle then changes the direction and moves towards +x and reaches x = 4 at t = 3 sec.
Total distance travelled = Distance travelled towards negative axis + Distance travelled towards positive axis
Total distance travelled = 1 + 4
Total distance travelled = 5m
#SPJ2