- An object moves along x-axis such that its position varying with time t is given as x = 4t - t^2(x is in metre and time t in second). The distance travelled by the object from t = 0 to t = 3 s is (1) 3 m (2) 5 m (3) 12 m (4) 21 m
Answers
Explanation:
I will consider each of the answer options below.
⇒
A: The magnitude of the x-coordinate of the particle is
4
m
.
I assume this refers to the position of the particle at
t
=
0
, since it travels along the x-axis only. We can substitute
0
into the given equation for position and see what we get for x.
x
(
t
)
=
(
4
t
−
t
2
−
4
)
⇒
x
(
0
)
=
(
4
(
0
)
−
(
0
)
2
−
4
)
⇒
x
=
−
4
Therefore, the initial position of the particle is
−
4
m
, and
|
−
4
|
=
4
, so answer A is correct.
⇒
B: The magnitude of the average velocity is equal to the average speed for the time interval
t
=
0
to
t
=
2
seconds.
We can compute the average velocity and average speed independently and compare the values we obtain.
The most important distinction between the two quantities is that the average speed is concerned with the distance the object travels over a given time period, whereas the average velocity is concerned with the displacement of the object over a given time period.
v
avg
=
Δ
x
Δ
t
We will first calculate the displacement
Δ
x
, where
Δ
x
=
x
f
−
x
i
.
x
i
=
x
(
0
)
=
−
4
m
x
f
=
x
(
2
)
=
0
m
Therefore,
Δ
x
=
0
−
(
−
4
)
=
4
m
.
We are given
Δ
t
=
t
f
−
t
i
=
2
−
0
=
2
s
v
avg
=
4
m
2
s
v
avg
=
2
m
/
s
(in the direction of the positive x-axis)
Now we calculate the average speed:
s
avg
=
Δ
d
Δ
t
At
t
=
0
, the particle is at
x
(
0
)
=
−
4
m
At
t
=
1
, the particle is at
x
(
1
)
=
−
1
m
At
t
=
2
, the particle is at
x
(
2
)
=
0
m
So, the object traveled a total distance of
(
3
+
1
)
m
=
4
m
Therefore,
Δ
d
=
4
m
and the time period is still
2
s
s
avg
=
4
m
2
s