Math Problem Statement

The position of a car is given by the equation s(t) = 5t - 5t^2. Find: (a) The displacement of the car between t = 3s and t = 5s. (b) The average velocity during the same time interval. (c) The instantaneous velocity at t = 3s and t = 5s.

Solution

The problem involves the position of a car given by the equation s(t)=5t5t2s(t) = 5t - 5t^2, and it has three parts:

  1. Find the displacement of the car between t=3st = 3 \, \text{s} and t=5st = 5 \, \text{s}.
  2. Determine the average velocity during the same time interval.
  3. Find the instantaneous velocity at t=3st = 3 \, \text{s} and t=5st = 5 \, \text{s}.

Let's solve each part step by step:


Solution

1. Displacement of the car

The displacement is the change in position of the car between t1=3t_1 = 3 and t2=5t_2 = 5.
The position function is:
s(t)=5t5t2s(t) = 5t - 5t^2

Evaluate s(t)s(t) at t=3t = 3 and t=5t = 5:
s(3)=5(3)5(32)=1545=30ms(3) = 5(3) - 5(3^2) = 15 - 45 = -30 \, \text{m}
s(5)=5(5)5(52)=25125=100ms(5) = 5(5) - 5(5^2) = 25 - 125 = -100 \, \text{m}

The displacement is:
Δs=s(5)s(3)=100(30)=100+30=70m\Delta s = s(5) - s(3) = -100 - (-30) = -100 + 30 = -70 \, \text{m}

So, the displacement is 70m-70 \, \text{m}.


2. Average velocity

The average velocity is given by:
vavg=ΔsΔtv_{\text{avg}} = \frac{\Delta s}{\Delta t}
Here, Δt=t2t1=53=2s\Delta t = t_2 - t_1 = 5 - 3 = 2 \, \text{s}, and Δs=70m\Delta s = -70 \, \text{m}.

vavg=702=35m/sv_{\text{avg}} = \frac{-70}{2} = -35 \, \text{m/s}

The average velocity is 35m/s-35 \, \text{m/s}.


3. Instantaneous velocity

The instantaneous velocity is the derivative of s(t)s(t):
v(t)=ddt(5t5t2)=510tv(t) = \frac{d}{dt} \left( 5t - 5t^2 \right) = 5 - 10t

Now evaluate v(t)v(t) at t=3t = 3 and t=5t = 5:
v(3)=510(3)=530=25m/sv(3) = 5 - 10(3) = 5 - 30 = -25 \, \text{m/s}
v(5)=510(5)=550=45m/sv(5) = 5 - 10(5) = 5 - 50 = -45 \, \text{m/s}

The instantaneous velocities are:

  • At t=3t = 3: 25m/s-25 \, \text{m/s}
  • At t=5t = 5: 45m/s-45 \, \text{m/s}

Final Answers

  1. Displacement: 70m-70 \, \text{m}
  2. Average velocity: 35m/s-35 \, \text{m/s}
  3. Instantaneous velocity:
    • At t=3t = 3: 25m/s-25 \, \text{m/s}
    • At t=5t = 5: 45m/s-45 \, \text{m/s}

Would you like detailed steps for any part or further clarifications?


5 Related Questions

  1. How is displacement different from total distance traveled?
  2. Can the average velocity ever equal the instantaneous velocity in any scenario?
  3. What does a negative velocity indicate in the context of motion?
  4. How would the calculations change if the equation included a constant term?
  5. What is the significance of the slope of the s(t)s(t) graph in interpreting velocity?

Tip

Always check the units of your answers to ensure consistency and correctness in physics problems.

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Math Problem Analysis

Mathematical Concepts

Algebra
Calculus
Kinematics

Formulas

Displacement formula: Δs = s(t2) - s(t1)
Average velocity formula: v_avg = Δs / Δt
Instantaneous velocity: v(t) = ds(t)/dt

Theorems

Derivative as a rate of change

Suitable Grade Level

Grades 11-12