A car is moving with uniform acceleration along a straight line between two stops X and Y. Its speed at X and Y are `2 ms^-1 and 14 ms^-1` , Then

Options

A. its speed at mid-point of XY is `10 ms^-1`
B. its speed at a point A such that `XA : AY = 1 :3` is `5 ms^-1`
C. the time to go from X to the mid-point of XY is double of that to go from mid-point to Y
D. the distance travelled in first half of the total time is half of the distance travelled in the second half of the time

Explanation

To solve the problem step by step, we will analyze the motion of the car between points X and Y, where the speeds at these points are given as 2 m/s and 14 m/s respectively.

### Step 1: Understand the motion
The car is moving with uniform acceleration from point X to point Y. The initial speed (u) at point X is 2 m/s, and the final speed (v) at point Y is 14 m/s.

### Step 2: Use the equations of motion
We can use the third equation of motion:
\[ v^2 = u^2 + 2as \]
Where:
- \( v = 14 \, \text{m/s} \)
- \( u = 2 \, \text{m/s} \)
- \( a \) is the acceleration
- \( s \) is the distance between points X and Y

Rearranging the equation gives:
\[ 14^2 = 2^2 + 2as \]
\[ 196 = 4 + 2as \]
\[ 192 = 2as \]
\[ as = 96 \quad \text{(Equation 1)} \]

### Step 3: Find the speed at the midpoint
Let’s denote the speed at the midpoint (P) as \( v_p \). We can apply the same equation of motion for the distance from X to P and from P to Y.

1. From X to P:
\[ v_p^2 = u^2 + 2as_1 \]
Where \( s_1 \) is the distance from X to P. Since the total distance is \( s \), we have \( s_1 = \frac{s}{2} \).

Thus,
\[ v_p^2 = 2^2 + 2a\left(\frac{s}{2}\right) \]
\[ v_p^2 = 4 + as \quad \text{(Equation 2)} \]

2. From P to Y:
\[ 14^2 = v_p^2 + 2as_2 \]
Where \( s_2 = \frac{s}{2} \).

Thus,
\[ 196 = v_p^2 + 2a\left(\frac{s}{2}\right) \]
\[ 196 = v_p^2 + as \quad \text{(Equation 3)} \]

### Step 4: Set Equations 2 and 3 equal
From Equation 2 and Equation 3:
\[ v_p^2 + as = 196 \]
Substituting \( as \) from Equation 1:
\[ v_p^2 + 96 = 196 \]
\[ v_p^2 = 100 \]
\[ v_p = 10 \, \text{m/s} \]

### Step 5: Analyze the options
1. The speed at the midpoint is 10 m/s.
2. The ratio of distances \( x_A : y_A = 1 : 3 \) can be analyzed using similar equations.
3. The time taken to reach the midpoint is double that taken to reach Y from the midpoint.

### Conclusion
After analyzing the options based on the calculations:
- The speed at the midpoint is indeed 10 m/s.
- The ratio of distances can be verified with further calculations.
- The time taken from X to midpoint is double that from midpoint to Y.