Math Problem Statement

A bike rider pedals with constant acceleration to reach a velocity of 7,5 m/s over a time of 4.5 s. During the period of acceleration, the bike's displacement is 19 m. What was the initial velocity of the bike?

Solution

We can solve this problem using the kinematic equations for motion with constant acceleration. The equation that relates displacement (dd), initial velocity (v0v_0), time (tt), and acceleration (aa) is:

d=v0t+12at2d = v_0 t + \frac{1}{2} a t^2

We also know that acceleration can be found using the following equation, given the final velocity (vv) and the initial velocity (v0v_0):

v=v0+atv = v_0 + a t

Given:

  • Final velocity (vv) = 7.5 m/s
  • Time (tt) = 4.5 s
  • Displacement (dd) = 19 m

We have two unknowns, v0v_0 (initial velocity) and aa (acceleration). To find v0v_0, we'll first find aa using the second equation, then substitute it into the first equation.

Step 1: Find acceleration (aa)

Rearrange the second equation to solve for aa:

a=vv0ta = \frac{v - v_0}{t}

Substitute this into the first equation.

Step 2: Substitute aa into the displacement equation

d=v0t+12(vv0t)t2d = v_0 t + \frac{1}{2} \left(\frac{v - v_0}{t}\right) t^2

Simplify this equation:

d=v0t+12(vv0)td = v_0 t + \frac{1}{2} (v - v_0) t

d=v0t+12vt12v0td = v_0 t + \frac{1}{2} v t - \frac{1}{2} v_0 t

d=(v012v0)t+12vtd = \left(v_0 - \frac{1}{2} v_0\right) t + \frac{1}{2} v t

d=12v0t+12vtd = \frac{1}{2} v_0 t + \frac{1}{2} v t

Multiply both sides by 2 to eliminate the fractions:

2d=v0t+vt2d = v_0 t + v t

Now, solve for v0v_0:

v0t=2dvtv_0 t = 2d - v t

v0=2dvttv_0 = \frac{2d - v t}{t}

Step 3: Substitute the known values

v0=2(19)(7.5)(4.5)4.5v_0 = \frac{2(19) - (7.5)(4.5)}{4.5}

Let's calculate this.

v0=3833.754.5=4.254.5v_0 = \frac{38 - 33.75}{4.5} = \frac{4.25}{4.5}

v00.94m/sv_0 \approx 0.94 \, \text{m/s}

Conclusion:

The initial velocity of the bike was approximately 0.94 m/s.

Would you like more details or explanations?

Here are some related questions you might find helpful:

  1. How do we calculate acceleration when displacement is not given?
  2. What if the rider starts from rest—how would the equations change?
  3. How does constant acceleration affect the total displacement?
  4. Can we calculate the rider's average velocity during this period?
  5. What other kinematic equations can be used for constant acceleration?

Tip: Always check if you can simplify your problem using known values before solving for unknowns in more complex equations.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Kinematics
Linear Motion
Constant Acceleration

Formulas

d = v_0 t + (1/2) a t^2
v = v_0 + a t
a = (v - v_0) / t
v_0 = (2d - v t) / t

Theorems

Kinematic equations of motion

Suitable Grade Level

Grades 10-12