For an object moving with uniform acceleration, travelling 50m in 5th sec, 70m in 7th sec.
(a)Its initial velocity is 5 m/s
(b) Its acceleration is `20 "m/s"^2`
(c)Its travels 100 m in 9th sec
(d) Its average velocity during 9th sec is 90 m/s

Options

A. a is true
B. a & d are true
C. a & b are true
D. a & c are true

Answer

To solve the problem step by step, we will use the equations of motion for uniformly accelerated motion.

### Step 1: Understand the Displacement in nth Second Formula
The displacement \( s_n \) during the nth second can be calculated using the formula:
\[ s_n = u + \frac{a}{2} (2n - 1) \]
where:
- \( s_n \) is the displacement in the nth second,
- \( u \) is the initial velocity,
- \( a \) is the acceleration,
- \( n \) is the second during which the displacement is measured.

### Step 2: Calculate Displacement in the 5th Second
Given that the displacement in the 5th second is 50 m:
\[ s_5 = u + \frac{a}{2} (2 \cdot 5 - 1) = 50 \]
This simplifies to:
\[ u + \frac{a}{2} \cdot 9 = 50 \]
or
\[ u + 4.5a = 50 \quad \text{(Equation 1)} \]

### Step 3: Calculate Displacement in the 7th Second
Given that the displacement in the 7th second is 70 m:
\[ s_7 = u + \frac{a}{2} (2 \cdot 7 - 1) = 70 \]
This simplifies to:
\[ u + \frac{a}{2} \cdot 13 = 70 \]
or
\[ u + 6.5a = 70 \quad \text{(Equation 2)} \]

### Step 4: Solve the Equations Simultaneously
Now we have two equations:
1. \( u + 4.5a = 50 \)
2. \( u + 6.5a = 70 \)

Subtract Equation 1 from Equation 2:
\[ (u + 6.5a) - (u + 4.5a) = 70 - 50 \]
This simplifies to:
\[ 2a = 20 \]
Thus,
\[ a = 10 \, \text{m/s}^2 \]

### Step 5: Substitute \( a \) Back to Find \( u \)
Now substitute \( a = 10 \) back into Equation 1:
\[ u + 4.5 \cdot 10 = 50 \]
This simplifies to:
\[ u + 45 = 50 \]
Thus,
\[ u = 5 \, \text{m/s} \]

### Step 6: Calculate Displacement in the 9th Second
Now, we calculate the displacement in the 9th second:
\[ s_9 = u + \frac{a}{2} (2 \cdot 9 - 1) \]
Substituting \( u = 5 \) and \( a = 10 \):
\[ s_9 = 5 + \frac{10}{2} \cdot 17 \]
This simplifies to:
\[ s_9 = 5 + 5 \cdot 17 = 5 + 85 = 90 \, \text{m} \]

### Step 7: Calculate Average Velocity in the 9th Second
The average velocity \( v_{avg} \) during the 9th second is given by:
\[ v_{avg} = \frac{s_9}{1} = 90 \, \text{m/s} \]

### Summary of Results
- Initial velocity \( u = 5 \, \text{m/s} \) (Option a is correct)
- Acceleration \( a = 10 \, \text{m/s}^2 \) (Option b is incorrect)
- Displacement in the 9th second is 90 m (Option c is correct)
- Average velocity during the 9th second is 90 m/s (Option d is correct)

### Final Answer
Correct options are (a), (c), and (d).