Math Problem Statement

body of mass 10 covers 20 m in its 2nd second and 50 m in its 3rd second of its journey. Compute its initial velocity and acceleration.

Solution

To solve this problem, we'll use the equations of motion for uniformly accelerated linear motion. Let's define the variables first:

  • uu: initial velocity (m/s)
  • aa: acceleration (m/s²)
  • sns_n: distance covered in the nnth second
  • tnt_n: time at the nnth second

Given:

  • Mass of the body, m=10m = 10 kg (though not needed to solve this problem)
  • Distance covered in the 2nd second, s2=20s_2 = 20 m
  • Distance covered in the 3rd second, s3=50s_3 = 50 m

Equations to use:

The distance covered in the nnth second is given by: sn=u+a2(2n1)s_n = u + \frac{a}{2} \cdot (2n - 1)

Step 1: Using the equation for the 2nd second

For n=2n = 2: s2=u+a2(2×21)=u+3a2s_2 = u + \frac{a}{2} \cdot (2 \times 2 - 1) = u + \frac{3a}{2} Given s2=20s_2 = 20 m, we have: 20=u+3a2(Equation 1)20 = u + \frac{3a}{2} \quad \text{(Equation 1)}

Step 2: Using the equation for the 3rd second

For n=3n = 3: s3=u+a2(2×31)=u+5a2s_3 = u + \frac{a}{2} \cdot (2 \times 3 - 1) = u + \frac{5a}{2} Given s3=50s_3 = 50 m, we have: 50=u+5a2(Equation 2)50 = u + \frac{5a}{2} \quad \text{(Equation 2)}

Step 3: Solving the equations

Subtract Equation 1 from Equation 2: 5020=(u+5a2)(u+3a2)50 - 20 = \left(u + \frac{5a}{2}\right) - \left(u + \frac{3a}{2}\right) 30=5a23a2=2a2=a30 = \frac{5a}{2} - \frac{3a}{2} = \frac{2a}{2} = a

So, a=30a = 30 m/s².

Now, substitute a=30a = 30 m/s² back into Equation 1: 20=u+3×30220 = u + \frac{3 \times 30}{2} 20=u+4520 = u + 45 u=2045=25 m/su = 20 - 45 = -25 \text{ m/s}

Final Answers:

  • Initial velocity u=25u = -25 m/s (the negative sign indicates the direction is opposite to the chosen positive direction).
  • Acceleration a=30a = 30 m/s².

Would you like further details on any part of this solution or have any other questions?

Related Questions:

  1. How would the time of travel change if the acceleration was halved?
  2. What would be the final velocity after 5 seconds of motion?
  3. How does the mass of the body affect the acceleration and velocity in this context?
  4. What would be the displacement after the first 3 seconds?
  5. How would the problem change if there was a constant opposing force?

Tip:

Always check the direction of motion when working with initial velocities and accelerations, especially if the initial velocity is negative, as it can affect the interpretation of the results.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Kinematics
Uniformly Accelerated Motion

Formulas

Equations of motion for uniformly accelerated linear motion

Theorems

-

Suitable Grade Level

High School