A body moving with uniform acceleration describes 40 m in the first 5 s and 70 m in the next 5 s.its initial velocity will be

Options

A. (a)`4 ms^(-1)`
B. (b)`2.5 ms^(-1)`
C. (c)`5 ms^(-1)`
D. (d)`11 ms^(-1)`

Answer

To solve the problem step by step, we will use the equations of motion for a body moving with uniform acceleration.

### Step 1: Understand the problem
We know that a body covers 40 m in the first 5 seconds and 70 m in the next 5 seconds. We need to find the initial velocity (u) of the body.

### Step 2: Use the equations of motion
We can use the equation of motion for the first 5 seconds:
\[ S_1 = ut + \frac{1}{2} a t^2 \]
Where:
- \( S_1 = 40 \, \text{m} \) (distance covered in the first 5 seconds)
- \( t_1 = 5 \, \text{s} \)

Substituting the values:
\[ 40 = u(5) + \frac{1}{2} a (5^2) \]
This simplifies to:
\[ 40 = 5u + \frac{25}{2} a \] (Equation 1)

### Step 3: Use the equations of motion for the next 5 seconds
For the next 5 seconds (from 5s to 10s), the total distance covered in 10 seconds is:
\[ S_2 = ut + \frac{1}{2} a t^2 \]
Where:
- \( S_2 = 40 + 70 = 110 \, \text{m} \) (total distance covered in 10 seconds)
- \( t_2 = 10 \, \text{s} \)

Substituting the values:
\[ 110 = u(10) + \frac{1}{2} a (10^2) \]
This simplifies to:
\[ 110 = 10u + 50a \] (Equation 2)

### Step 4: Solve the equations simultaneously
Now we have two equations:
1. \( 40 = 5u + \frac{25}{2} a \)
2. \( 110 = 10u + 50a \)

From Equation 1, we can express \( a \) in terms of \( u \):
\[ \frac{25}{2} a = 40 - 5u \]
\[ a = \frac{80 - 10u}{25} \] (Equation 3)

Now substitute Equation 3 into Equation 2:
\[ 110 = 10u + 50\left(\frac{80 - 10u}{25}\right) \]
\[ 110 = 10u + 2(80 - 10u) \]
\[ 110 = 10u + 160 - 20u \]
\[ 110 = 160 - 10u \]
\[ 10u = 160 - 110 \]
\[ 10u = 50 \]
\[ u = 5 \, \text{m/s} \]

### Final Answer
The initial velocity \( u \) of the body is \( 5 \, \text{m/s} \).

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