Need Help with 1D Kinematics Problems? Here's Step-by-Step Guidance!

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In summary, two problems were discussed. The first one involved a sports car moving at a constant speed and then braking to a stop. The acceleration of the car was asked in terms of g's, given that the car traveled 115 m in 5.3 s and stopped in 4.0 s. In the second problem, a rock was dropped from a sea cliff and the sound of it striking the ocean was heard 4.2 s later. The question asked for the height of the cliff, given the speed of sound was 340 m/s. No information was provided about the speed of the car in the first problem, so it cannot be solved.
  • #1
AbbyGirl
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Hi Guys! I'm having trouble with 2 problems that I only have 1 attempt left to answer correctly so if someone could please just tell me step by step how to complete them that would be great. Thanks to those who help.

Problem 1: A sports car moving at constant speed travels 115 m in 5.3 s. If it then brakes and comes to a stop in 4.0 s, what is its acceleration? Express the answer in terms of g's, where g = 9.80 m/s2. (Take the positive direction to be the direction of travel.)

Problem 2:A rock is dropped from a sea cliff and the sound of it striking the ocean is heard 4.2 s later. If the speed of sound is 340 m/s, how high is the cliff?

Again, thanks a lot.

-Abby
 
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  • #2
What have you tried so far?

In the first one how fast was the car going? You need that to figure out how fast it decelerates.

What about the second one?
 
  • #3


For problem 1, we can use the formula a = (vf - vi)/t, where a is acceleration, vf is final velocity, vi is initial velocity, and t is time. In this case, the car starts at a constant speed, so vi = vf. We can also use the formula d = vit + 1/2at^2, where d is distance, vi is initial velocity, a is acceleration, and t is time. We know that the car travels 115 m in 5.3 s, so we can set up the equation: 115 = (vf)(5.3) + 1/2a(5.3)^2. Since vf = vi, we can simplify this to 115 = vi(5.3) + 1/2a(5.3)^2. We also know that the car comes to a stop in 4.0 s, so we can set up another equation: 0 = vi(4.0) + 1/2a(4.0)^2. Now we have two equations with two unknowns (vi and a), so we can solve for them using algebra. Once we have the values for vi and a, we can express the acceleration in terms of g's by dividing the value by 9.80 m/s^2.

For problem 2, we can use the formula d = vit + 1/2at^2, where d is distance, vi is initial velocity, a is acceleration, and t is time. In this case, we know that the rock is dropped from rest (vi = 0), so the equation becomes d = 1/2at^2. We also know that the sound of the rock hitting the ocean is heard 4.2 s later, so we can plug that value in for t. We also know that the speed of sound is 340 m/s, so we can plug that value in for a. Now we have one equation with one unknown (d), so we can solve for the height of the cliff using algebra.

I hope this helps! Remember to always show your work and units in your answers. Good luck!
 

1. What is 1 Dimensional Kinematics?

1 Dimensional Kinematics is the study of motion in one dimension, typically along a straight line. It involves the analysis of position, velocity, and acceleration of objects.

2. What is displacement in 1 Dimensional Kinematics?

Displacement is the change in position of an object in a specific direction. It is calculated by subtracting the initial position from the final position.

3. How is velocity calculated in 1 Dimensional Kinematics?

Velocity is calculated by dividing the change in displacement by the change in time. It is a vector quantity that includes both magnitude and direction.

4. What is acceleration in 1 Dimensional Kinematics?

Acceleration is the rate of change of an object's velocity. It is calculated by dividing the change in velocity by the change in time. It is also a vector quantity.

5. What are the equations used in 1 Dimensional Kinematics?

The main equations used in 1 Dimensional Kinematics are:

  • Velocity: v = Δx/Δt
  • Acceleration: a = Δv/Δt
  • Displacement: Δx = v0t + 1/2at2
  • Final velocity: v = v0 + at

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