A car moving with constant acceleration covers the distance between two points 180 m apart in 6 sec. Its speed as it passes the second point is 45 m/s. What is its acceleration and its speed at the first point

Options

A. `-5 "m/s"^2 , 15 "m/s"`
B. `-15 "m/s"^2 , 5 "m/s"`
C. `-5 "m/s"^2 , -15 "m/s"`
D. `5 "m/s"^2 , 15 "m/s"`

Answer

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

### Given Data:
- Distance (s) = 180 m
- Time (t) = 6 s
- Final speed (v) = 45 m/s
- Initial speed (u) = ? (to be determined)
- Acceleration (a) = ? (to be determined)

### Step 1: Use the equation of motion to find the initial speed (u).
We can use the equation for distance in uniformly accelerated motion:
\[ s = \frac{(u + v)}{2} \cdot t \]
Substituting the known values:
\[ 180 = \frac{(u + 45)}{2} \cdot 6 \]
Now, simplify the equation:
\[ 180 = 3(u + 45) \]
\[ 180 = 3u + 135 \]
Subtract 135 from both sides:
\[ 45 = 3u \]
Now, divide by 3:
\[ u = 15 \, \text{m/s} \]

### Step 2: Now, find the acceleration (a).
We can use the formula for acceleration:
\[ a = \frac{(v - u)}{t} \]
Substituting the known values:
\[ a = \frac{(45 - 15)}{6} \]
\[ a = \frac{30}{6} = 5 \, \text{m/s}^2 \]

### Final Answers:
- Initial speed (u) = 15 m/s
- Acceleration (a) = 5 m/s²