An object moves with constant acceleration 3.65 m/s2 and over a time interval reaches a final velocity of 10.4 m/s.
(a) If its initial velocity is 5.2 m/s, what is its displacement during the time interval?
m
(b) What is the distance it travels during this interval?
m
(c) If its initial velocity is
−5.2
m/s, what is its displacement during the time interval?
m
(d) What is the total distance it travels during the interval in part (c)?
m
To answer these questions, we can use the equations of motion for an object with constant acceleration. The four equations are:
1. v = u + at
2. s = ut + 0.5at^2
3. v^2 = u^2 + 2as
4. s = ((u + v)/2)t
Where:
- s is the displacement
- u is the initial velocity
- v is the final velocity
- a is the acceleration
- t is the time interval
(a) To find the displacement when the initial velocity is 5.2 m/s and the final velocity is 10.4 m/s:
Given:
u = 5.2 m/s
v = 10.4 m/s
a = 3.65 m/s^2
We need to find s.
We can use equation (3):
v^2 = u^2 + 2as
Plug in the values:
(10.4 m/s)^2 = (5.2 m/s)^2 + 2(3.65 m/s^2)s
Solve for s:
107.84 m^2/s^2 - 27.04 m^2/s^2 = 7.3 m/s^2 * s
80.8 m^2/s^2 = 7.3 m/s^2 * s
s ≈ 11.1 m
Therefore, the displacement during the time interval is 11.1 m.
(b) To find the distance traveled during this interval:
Distance is the total path length traveled, regardless of direction. Since acceleration can be positive or negative, the object may change direction during motion.
To find the distance, we can use the equation:
distance = |s|
In this case, the displacement is positive, so the distance traveled is also equal to the displacement:
distance = 11.1 m
Therefore, the distance traveled during this interval is 11.1 m.
(c) To find the displacement when the initial velocity is -5.2 m/s:
Given:
u = -5.2 m/s
v = 10.4 m/s
a = 3.65 m/s^2
We need to find s.
Using equation (3):
(10.4 m/s)^2 = (-5.2 m/s)^2 + 2(3.65 m/s^2)s
Solve for s:
107.84 m^2/s^2 = 27.04 m^2/s^2 + 7.3 m/s^2 * s
80.8 m^2/s^2 = 7.3 m/s^2 * s
s ≈ 11.1 m
Therefore, the displacement during the time interval is 11.1 m.
(d) To find the total distance traveled during this interval:
Since the initial velocity is negative (-5.2 m/s), the object changes direction during motion. The total distance is the sum of the distances traveled in each direction.
In this case, the distance traveled when the initial velocity is -5.2 m/s is equal to 11.1 m. Since it travels the same distance in the opposite direction, the total distance is:
total distance = 11.1 m + 11.1 m = 22.2 m
Therefore, the total distance traveled during the interval in part (c) is 22.2 m.
To solve this problem, we can use the equations of motion for an object with constant acceleration.
(a) To find the displacement (Δx) during the time interval, we can use the equation:
v^2 = u^2 + 2aΔx
where:
v = final velocity = 10.4 m/s
u = initial velocity = 5.2 m/s
a = acceleration = 3.65 m/s^2
Rearranging the equation to solve for Δx:
Δx = (v^2 - u^2) / (2a)
Plugging in the given values:
Δx = (10.4^2 - 5.2^2) / (2 * 3.65)
= (108.16 - 27.04) / 7.3
= 81.12 / 7.3
≈ 11.12 m
Therefore, the displacement during the time interval is approximately 11.12 m.
(b) The distance traveled can be calculated by taking the absolute value of the displacement. Therefore, the distance traveled is also approximately 11.12 m.
(c) To find the displacement with an initial velocity of -5.2 m/s, we can use the same equation as in part (a), but with a negative initial velocity:
Δx = (v^2 - u^2) / (2a)
Plugging in the given values:
Δx = (10.4^2 - (-5.2)^2) / (2 * 3.65)
= (108.16 - 27.04) / 7.3
= 81.12 / 7.3
≈ 11.12 m
Therefore, the displacement during the time interval with an initial velocity of -5.2 m/s is also approximately 11.12 m.
(d) To find the total distance traveled during the interval in part (c), we take the absolute value of the displacement, just like in part (b). Therefore, the total distance traveled is also approximately 11.12 m.
Assume that positive x-axis directed to the right
(a)(b) v=vₒ+at
t=( v-vₒ)/a=(10.4-5.2)/3.65=1.42 s
displacement = distance= vₒt+at²/2=5.2•1.42+3.65•1.42²/2=
7.384+3.68 =11.064 m
(from origin point - to theright)
(c) v=-vₒ+at
t1=(0+vₒ)/a=5.2/3.65=1.42 s.
s1 =vₒ²/2a=5.2²/2•3.65=3.7 m (to the left)
v=a•t2
t2=v/a=10.4/3.65=2.89 s.
s2= at²/2=3.65•2.89²/2=15.23 m (to the right)
Distance=3.7 +15.23=18.93 m
Displacement =15.23-3.7=11.53 m