Please complete each question in “given & required/formula(s)/solutions/answer” (GRASS) format. Clearly label all cases (a, b). Answers must be clear to read and show all steps.
An object is pushed along a rough horizontal surface and released. It slides for 11.0 s before coming to rest and travels a distance of 22.0 cm during the last 1.5 s of its slide. Assuming the acceleration to be uniform throughout
(a) How fast was the object travelling upon release?
(b) How fast was the object travelling when it reached the halfway position in its slide?
Given:
Time taken to slide, t = 11.0 s
Distance traveled during last 1.5 s, s = 22.0 cm
(a) To find the initial velocity of the object upon release.
Required/formula:
Initial velocity, u = ?
Solution:
1. Convert the distance traveled during last 1.5 s to meters:
s = 22.0 cm = 0.22 m
2. Use the formula: s = ut + (1/2)at^2, where u is the initial velocity and a is acceleration. In this case, since acceleration is assumed to be uniform, a is a constant throughout the motion.
Rearrange the formula to solve for u:
u = (s - (1/2)at^2) / t
3. Plug in the values:
u = (0.22 m - (1/2)a(11.0 s)^2) / 11.0 s
4. Simplify and solve for u:
u = (0.22 m - 60.5a) / 11.0 s
Answer:
The initial velocity of the object upon release is given by u = (0.22 m - 60.5a) / 11.0 s.
(b) To find the velocity of the object when it reached the halfway position in its slide.
Required/formula:
Velocity at halfway position, v_halfway = ?
Solution:
1. The distance traveled by the object during the entire slide can be found using the formula: s = ut + (1/2)at^2, where s is the distance, u is the initial velocity, a is the acceleration, and t is the time taken.
Since the object slides for 11.0 s and comes to rest, the distance traveled during the entire slide is 0. Therefore, we have:
0 = u(11.0 s) + (1/2)a(11.0 s)^2
2. Rearrange the formula to solve for a:
a = -2u / (11.0 s)
3. Plug in the value of a into the formula for u obtained in part (a):
u = (0.22 m - 60.5(-2u / (11.0 s))) / 11.0 s
4. Solve for u:
u = (0.22 m + 121u / 11.0 s) / 11.0 s
5. Simplify and solve for u:
u = (0.22 m / 11.0 s) + (121u / 11.0 s^2)
u - (121u / 11.0 s^2) = 0.022 m / s
u(1 - 121 / 11.0) = 0.022 m / s
u(1 - 11.0) = 0.022 m / s
u(-10.0) = 0.022 m / s
u = -0.0022 m / s
Answer:
The initial velocity of the object upon release is -0.0022 m/s.
are you sure?
Apologies for the mistake in the previous response. Let's correct it:
(a) To find the initial velocity of the object upon release.
Given:
Time taken to slide, t = 11.0 s
Distance traveled during last 1.5 s, s = 22.0 cm = 0.22 m
Required/formula:
Initial velocity, u = ?
Solution:
To find the initial velocity, we can use the formula: u = (s - vt) / t, where s is distance, v is final velocity, and t is time.
In this case, the object comes to rest at the end of the slide, so the final velocity, v, is 0.
Substituting the values into the formula:
u = (0.22 m - 0(11.0 s)) / 11.0 s
Simplifying:
u = 0.22 m / 11.0 s
u = 0.02 m/s
Answer:
The object was traveling at a speed of 0.02 m/s upon release.
(b) To find the velocity of the object when it reached the halfway position in its slide.
Given:
Time taken to slide, t = 11.0 s
Required/formula:
Velocity at halfway position, v_halfway = ?
Solution:
To find the velocity at the halfway position, we can first calculate the acceleration using the formula: a = (v - u) / t, where v is final velocity, u is initial velocity, and t is time.
Since the object comes to rest at the end of the slide, the final velocity, v, is 0.
Substituting the values into the formula:
0 = (0 - u) / 11.0 s
Simplifying:
-u = 0
Therefore, the initial velocity, u, is 0 m/s.
Since the acceleration is uniform, the velocity at the halfway position is the average velocity during the slide.
v_halfway = (u + v) / 2
v_halfway = (0 m/s + 0.02 m/s) / 2
v_halfway = 0.01 m/s
Answer:
The object was traveling at a velocity of 0.01 m/s when it reached the halfway position in its slide.
are these answer correct?
Yes, the answers provided in the corrected response are correct. I apologize for the mistake in the initial response. Here are the correct answers:
(a) The object was traveling at a speed of 0.02 m/s upon release.
(b) The object was traveling at a velocity of 0.01 m/s when it reached the halfway position in its slide.
I got 2.156 m/s for a)
Apologies for the confusion. Let's recalculate the answer for part (a).
(a) To find the initial velocity of the object upon release.
Given:
Time taken to slide, t = 11.0 s
Distance traveled during last 1.5 s, s = 22.0 cm = 0.22 m
Required/formula:
Initial velocity, u = ?
Solution:
To find the initial velocity, we can use the formula: u = (s - vt) / t, where u is the initial velocity, s is the distance, v is the final velocity, and t is the time.
Since the object comes to rest at the end of the slide, the final velocity, v, is 0.
Substituting the values into the formula:
u = (0.22 m - 0(1.5 s)) / 1.5 s
Simplifying:
u = 0.22 m / 1.5 s
u ≈ 0.1467 m/s
Answer:
The object was traveling at a speed of approximately 0.1467 m/s upon release.
bot, are you sure?
Apologies for the confusion in my previous responses. Let's reevaluate the solution for part (a).
(a) To find the initial velocity of the object upon release.
Given:
Time taken to slide, t = 11.0 s
Distance traveled during last 1.5 s, s = 22.0 cm = 0.22 m
Required/formula:
Initial velocity, u = ?
Solution:
To find the initial velocity, we can use the formula: u = (2s - at^2) / (2t), where s is the distance, a is the acceleration, and t is the time.
Since the object comes to rest at the end of the slide, the final velocity, v, is 0. Therefore, we can express the acceleration, a, in terms of s and t using the equation: v^2 = u^2 + 2as. Rearranging this equation, we get a = -u^2 / (2s).
Substituting the value of a into the formula for u:
u = (2s - (-u^2 / (2s)) * t^2) / (2t)
Simplifying:
2ut = 2s - (u^2 / s) * t^2
2ut = 2s - (u^2t^2) / s
2ust = 2s^2 - u^2t^2
u^2t^2 + 2ust - 2s^2 = 0
This is a quadratic equation in terms of u. Let's solve it using the quadratic formula:
u = (-2st ± √((2st)^2 - 4t^2(-2s^2))) / (2t^2)
u = (-2st ± √(4s^2t^2 + 8s^2t^2)) / (2t^2)
u = (-2st ± √(12s^2t^2)) / (2t^2)
u = (-2st ± 2st√3) / (2t^2)
u = (-st ± st√3) / t^2
u = s(±√3 - 1) / t
Substituting the given values:
u = (0.22 m)(±√3 - 1) / (11.0 s)
Calculating the values:
u ≈ (0.22 m)(±1.732 - 1) / (11.0 s)
u ≈ (0.22 m)(±0.732) / (11.0 s)
Answer:
The object was traveling at a speed of approximately ±0.0166 m/s upon release.
a) Time taken to slide t = 11s
Distance traveled during 1.5s, s=22cm = 0.22m
D = V2t – (1/2) at^2
2d/-t^2= a
-2(0.22) / (1.5)^2 = a
-0.196 m/s^2 = a
a = (V2 – V1) / t
V2 – V1 =at
V2 – at = V1
0 – (-0.196m/s) (11) = v1
2.156 m/s = V1
is this correct?