Math Problem Statement

a car increases its speed at constant rate from 64 kph at A to 96 kph at 8. what is the magnitude of its acceleration 2 sec after the car passes point A? (14.3 m/s^2)

Solution

To find the magnitude of the car's acceleration 22 seconds after passing point AA, we can use the following steps.

Step 1: Convert the speeds from kph to m/s

First, let's convert the initial speed vA=64v_A = 64 kph and the final speed vB=96v_B = 96 kph into meters per second (m/s).

vA=64×10003600=17.78 m/sv_A = 64 \times \frac{1000}{3600} = 17.78 \text{ m/s} vB=96×10003600=26.67 m/sv_B = 96 \times \frac{1000}{3600} = 26.67 \text{ m/s}

Step 2: Determine the rate of acceleration

Since the acceleration is constant, we can use the formula:

a=ΔvΔta = \frac{\Delta v}{\Delta t}

Here, Δv=vBvA\Delta v = v_B - v_A, and Δt=8\Delta t = 8 seconds, which gives:

Δv=26.67m/s17.78m/s=8.89m/s\Delta v = 26.67 \, \text{m/s} - 17.78 \, \text{m/s} = 8.89 \, \text{m/s}

a=8.89m/s8s=1.11m/s2a = \frac{8.89 \, \text{m/s}}{8 \, \text{s}} = 1.11 \, \text{m/s}^2

Step 3: Find the speed after 2 seconds

The speed after 2 seconds, v2v_2, can be found using the equation:

v2=vA+a×tv_2 = v_A + a \times t

Here, t=2t = 2 seconds:

v2=17.78m/s+1.11m/s2×2s=19.00m/sv_2 = 17.78 \, \text{m/s} + 1.11 \, \text{m/s}^2 \times 2 \, \text{s} = 19.00 \, \text{m/s}

Step 4: Calculate the acceleration 2 seconds after point AA

The magnitude of the acceleration at 2 seconds is still the constant acceleration we calculated:

a=1.11m/s2a = 1.11 \, \text{m/s}^2

However, it seems you mentioned 14.3m/s214.3 \, \text{m/s}^2 in your question. This value doesn't match the result from the calculations. If there was a different interpretation or additional information regarding this problem, please clarify.

Would you like to go through the details or have any questions?


Related Questions:

  1. How would the problem change if the car's acceleration was not constant?
  2. What is the distance traveled by the car in the first 2 seconds?
  3. How would the acceleration change if the car reached 120120 kph in 1010 seconds?
  4. How does the unit conversion from kph to m/s impact the accuracy of the calculations?
  5. How would the scenario change if the car started from rest?

Tip: Always ensure you understand whether the problem assumes constant or variable acceleration, as this greatly influences the approach to solving it.

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Math Problem Analysis

Mathematical Concepts

Kinematics
Constant Acceleration

Formulas

Acceleration formula: a = (v_B - v_A) / t
Speed formula: v = v_A + a * t

Theorems

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Suitable Grade Level

High School