Math Problem Statement

If the car’s speed decreases at a constant rate from 80 mi/h to 50 mi/h in 3.0 s , what is the magnitude of its acceleration, assuming that it continues to move in a straight line?

Solution

To find the magnitude of the car's acceleration, we can use the formula for acceleration, which is defined as the change in velocity divided by the time taken:

a=ΔvΔta = \frac{\Delta v}{\Delta t}

Where:

  • aa is the acceleration,
  • Δv\Delta v is the change in velocity,
  • Δt\Delta t is the time over which the change occurs.

Step 1: Calculate the change in velocity

Δv=vfvi\Delta v = v_f - v_i Where:

  • vfv_f is the final velocity (50 mi/h),
  • viv_i is the initial velocity (80 mi/h).

Δv=50mi/h80mi/h=30mi/h\Delta v = 50 \, \text{mi/h} - 80 \, \text{mi/h} = -30 \, \text{mi/h}

Step 2: Convert velocity from miles per hour to feet per second

We need to convert the velocities from miles per hour (mi/h) to feet per second (ft/s) to maintain standard units for acceleration. The conversion factor is:

1mi/h=1.467ft/s1 \, \text{mi/h} = 1.467 \, \text{ft/s}

Thus: vi=80mi/h×1.467=117.36ft/sv_i = 80 \, \text{mi/h} \times 1.467 = 117.36 \, \text{ft/s} vf=50mi/h×1.467=73.35ft/sv_f = 50 \, \text{mi/h} \times 1.467 = 73.35 \, \text{ft/s}

Now, calculate the change in velocity in ft/s: Δv=73.35ft/s117.36ft/s=44.01ft/s\Delta v = 73.35 \, \text{ft/s} - 117.36 \, \text{ft/s} = -44.01 \, \text{ft/s}

Step 3: Calculate the acceleration

The time interval given is 3.0 seconds, so we can now find the acceleration:

a=44.01ft/s3.0s=14.67ft/s2a = \frac{-44.01 \, \text{ft/s}}{3.0 \, \text{s}} = -14.67 \, \text{ft/s}^2

Step 4: Find the magnitude of the acceleration

Since acceleration is a vector and we are asked for the magnitude, we take the absolute value:

a=14.67ft/s2|a| = 14.67 \, \text{ft/s}^2

Thus, the magnitude of the car's acceleration is 14.67 ft/s².

Would you like more details on this, or do you have any questions?

Here are some related questions you could explore:

  1. How would the acceleration change if the car slowed down over 5 seconds instead?
  2. What would be the car's speed after 2 seconds of deceleration?
  3. How would the acceleration be affected if the car started at a speed of 100 mi/h instead of 80 mi/h?
  4. What is the relationship between velocity and acceleration during constant deceleration?
  5. How would the acceleration change if the car came to a complete stop?

Tip: When converting units, always make sure to align the units properly before performing calculations to avoid errors.

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Math Problem Analysis

Mathematical Concepts

Kinematics
Linear Motion
Acceleration
Unit Conversion

Formulas

a = Δv / Δt
Δv = v_f - v_i
1 mi/h = 1.467 ft/s

Theorems

Basic principles of constant acceleration

Suitable Grade Level

Grades 9-12