are these answers correct?
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?
Time taken to slide t = 11s
Distance traveled during 1.5s, s=22cm = 0.22m
D = V2t – (1/2) at2
2d/-t
2= a
-2(0.22) / (1.5) 2 = a
-0.196 m/s2 = a
a = (V2 – V1) / t
V2 – V1 =at
V2 – at = V1
0 – (-0.196m/s) (11) = v1
2.156 m/s = V1
(b) How fast was the object travelling when it reached the halfway position in its slide?
Time taken to slide t = 11s
V1 = 2.156 m/s
T = 11s
V2 = 0
D =?
D = v1 t + ½ at2
D = (2.156m/s) (11s) + ½ (-0.196m/s2
) (10s)2
D = 13.92m
Halfway position = 11m
V1 = 2.156 m/s
A = -0.196 m/s2
2 ad = V22 – V12
V12 + 2ad = V22
(2.156 m/s) 2 + 2 (-0.196 m/s2
) (11m) = V22
0.336m/s = V22
√0.336m/s = V2
0.58m/s = V2
(a) Yes, the answer is correct. The object was traveling at 2.156 m/s upon release.
(b) No, the answer is not correct. The object was traveling at 0.58 m/s when it reached the halfway position in its slide.
(a) The initial velocity of the object, also known as the speed upon release, is calculated using the formula V2 - V1 = at, where V2 is the final velocity of 0 m/s, a is the acceleration of -0.196 m/s^2, and t is the time taken to slide of 11 s.
V2 - (-0.196 m/s^2) * 11 s = V1
0 m/s + (0.196 m/s^2 * 11 s) = V1
2.156 m/s = V1
Therefore, the object was traveling at a speed of 2.156 m/s upon release.
(b) The speed of the object when it reaches the halfway position is calculated using the formula V12 + 2ad = V22, where V1 is the initial velocity of 2.156 m/s, a is the acceleration of -0.196 m/s^2, and d is the distance traveled to reach the halfway position of 11 m.
V12 + 2(-0.196 m/s^2) * 11 m = V22
(2.156 m/s)^2 + 2(-0.196 m/s^2 * 11 m) = V22
0.336 m/s = V22
Taking the square root of both sides:
V2 = √(0.336 m/s)
V2 = 0.58 m/s
Therefore, the object was traveling at a speed of 0.58 m/s when it reached the halfway position in its slide.
To verify if these answers are correct, let's go through the calculations step by step.
(a) How fast was the object traveling upon release?
The time taken to slide, t = 11 s
Distance traveled during the last 1.5 s, s = 22 cm = 0.22 m
Using the equation of motion: D = V2t - (1/2)at^2
We can rearrange to solve for acceleration:
2d/-t = a
a = -2(0.22) / (1.5)^2
a = -0.196 m/s^2
Next, we can calculate the initial velocity, V1, using the equation: a = (V2 - V1) / t
V2 - at = V1
0 - (-0.196m/s) (11s) = V1
2.156 m/s = V1
Therefore, the object's initial velocity upon release is 2.156 m/s.
(b) How fast was the object traveling when it reached the halfway position in its slide?
Given:
V1 = 2.156 m/s
t = 11 s
V2 = 0
To find the distance traveled, D, we can use the equation: D = V1t + (1/2)at^2
D = (2.156 m/s) (11 s) + (1/2) (-0.196 m/s^2) (11 s)^2
D = 13.92 m
Since the halfway position is 11 m, we need to find the velocity when D = 11 m.
Using the equation of motion: V^2 = V1^2 + 2ad, where a = -0.196 m/s^2, and d = 11 m,
V^2 = (2.156 m/s)^2 + 2 (-0.196 m/s^2)(11 m)
V^2 = 0.336 m^2/s^2
Taking the square root, we find V = 0.58 m/s.
Therefore, the object's velocity when it reached the halfway position in its slide is 0.58 m/s.
To verify if these answers are correct, you can plug the values back into the equations and check if they match the given information.