Math Problem Statement

Given the ball was initially at rest, calculate the final velocity of the ball when it had traveled 10 meters using the provided acceleration vs distance graph.

Solution

To calculate the final velocity of the ball after it has traveled 10 meters, we use kinematics principles with the given graph.


Step 1: Understanding the graph

The graph shows a linear relationship between acceleration (m/s²) and distance (m).

From the graph:

  • At 0m0 \, \text{m}, acceleration a=0m/s2a = 0 \, \text{m/s}^2
  • At 25m25 \, \text{m}, acceleration a=15m/s2a = 15 \, \text{m/s}^2

Thus, the acceleration increases linearly as: a=kda = k \cdot d where dd is the distance and kk is the constant of proportionality.

From the graph: k=change in accelerationchange in distance=150250=0.6m/s2 per meter.k = \frac{\text{change in acceleration}}{\text{change in distance}} = \frac{15 - 0}{25 - 0} = 0.6 \, \text{m/s}^2 \text{ per meter}.

Therefore: a=0.6da = 0.6d


Step 2: Relating acceleration to velocity

The acceleration is not constant here; it depends on the distance. To find the velocity, we use the work-energy theorem (or integrate acceleration over distance).

The kinematic relationship between acceleration and velocity is: v2=v02+2adsv^2 = v_0^2 + 2 \int a \, ds Given:

  • v0=0m/sv_0 = 0 \, \text{m/s} (initial velocity),
  • a=0.6da = 0.6d.

Substitute a=0.6da = 0.6d into the integral: v2=0+20100.6ddsv^2 = 0 + 2 \int_0^{10} 0.6d \, ds


Step 3: Solving the integral

The integral of 0.6d0.6d with respect to dd is: 0.6dds=0.6d22=0.3d2\int 0.6d \, ds = 0.6 \cdot \frac{d^2}{2} = 0.3d^2

Now evaluate the integral from d=0d = 0 to d=10d = 10: 0100.6dds=0.3(10)20.3(0)2=0.3(100)=30\int_0^{10} 0.6d \, ds = 0.3(10)^2 - 0.3(0)^2 = 0.3(100) = 30

Thus: v2=230=60v^2 = 2 \cdot 30 = 60


Step 4: Calculating final velocity

Take the square root of v2v^2: v=607.75m/sv = \sqrt{60} \approx 7.75 \, \text{m/s}


Final Answer:

The final velocity of the ball after traveling 10 meters is approximately: 7.75m/s\boxed{7.75 \, \text{m/s}}


Do you want me to explain any part in more detail? Here are some related questions for further understanding:

  1. How do you derive a=0.6da = 0.6d from the graph?
  2. Why do we use the integral to calculate velocity when acceleration is not constant?
  3. How would the solution change if the acceleration were constant?
  4. What happens if the ball starts with an initial velocity other than 0m/s0 \, \text{m/s}?
  5. How can we plot the velocity vs distance graph for this motion?

Tip: When acceleration varies with position, integrating acceleration helps determine the velocity at any point.

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Math Problem Analysis

Mathematical Concepts

Kinematics
Integral Calculus
Linear Acceleration

Formulas

a = k * d (acceleration as a function of distance)
v^2 = v_0^2 + 2 * ∫ a ds (kinematic relation between velocity and acceleration)
∫ 0.6d ds = 0.3d^2 (integrating acceleration with respect to distance)

Theorems

Work-Energy Theorem
Kinematic Equations

Suitable Grade Level

Grades 10-12