At time `t = 0,` a car moving along a straight line has a velocity of `16 ms^-1.` It slows down with an acceleration of `-0.5t ms^-2,` where t is in second. Mark the correct statement (s).

Options

A. The direction of velocity changes at `t = 8 s`
B. The distance travelled in 4 s is approximately 58.67 m
C. The distance travelled by the particle in 10 s is 94 m
D. The speed of particle at `t = 10` s is `9 ms^-1`

Explanation

To solve the problem step by step, we need to analyze the motion of the car given its initial conditions and the acceleration function.

### Step 1: Understand the given information
- Initial velocity of the car, \( v_0 = 16 \, \text{m/s} \)
- Acceleration, \( a(t) = -0.5t \, \text{m/s}^2 \)

### Step 2: Relate acceleration to velocity
We know that acceleration is the rate of change of velocity:
\[ a = \frac{dv}{dt} \]
Thus, we can write:
\[ dv = a \, dt = -0.5t \, dt \]

### Step 3: Integrate to find velocity as a function of time
Integrating both sides:
\[ \int dv = \int -0.5t \, dt \]
This gives:
\[ v = -0.25t^2 + C \]
Where \( C \) is the constant of integration. We know that at \( t = 0 \), \( v = 16 \, \text{m/s} \):
\[ 16 = -0.25(0)^2 + C \implies C = 16 \]
Thus, the velocity function is:
\[ v(t) = 16 - 0.25t^2 \]

### Step 4: Determine when the velocity becomes zero
To find when the car stops (velocity becomes zero):
\[ 0 = 16 - 0.25t^2 \]
Solving for \( t \):
\[ 0.25t^2 = 16 \implies t^2 = 64 \implies t = 8 \, \text{s} \]
So, the direction of the velocity changes at \( t = 8 \, \text{s} \).

### Step 5: Find the distance traveled in 4 seconds
Using the velocity function to find the distance:
\[ s(t) = \int v(t) \, dt = \int (16 - 0.25t^2) \, dt \]
Integrating:
\[ s(t) = 16t - \frac{0.25t^3}{3} + C \]
At \( t = 0 \), \( s(0) = 0 \), so \( C = 0 \):
\[ s(t) = 16t - \frac{0.25t^3}{3} \]
Now, substituting \( t = 4 \):
\[ s(4) = 16(4) - \frac{0.25(4^3)}{3} = 64 - \frac{0.25(64)}{3} = 64 - \frac{16}{3} = 64 - 5.33 \approx 58.67 \, \text{m} \]

### Step 6: Find the distance traveled in 10 seconds
Substituting \( t = 10 \):
\[ s(10) = 16(10) - \frac{0.25(10^3)}{3} = 160 - \frac{0.25(1000)}{3} = 160 - \frac{250}{3} = 160 - 83.33 \approx 76.67 \, \text{m} \]

### Step 7: Calculate the total distance traveled until \( t = 10 \)
The distance traveled from \( t = 0 \) to \( t = 8 \) is \( s(8) \):
\[ s(8) = 16(8) - \frac{0.25(8^3)}{3} = 128 - \frac{0.25(512)}{3} = 128 - \frac{128}{3} = 128 - 42.67 \approx 85.33 \, \text{m} \]
After \( t = 8 \), the car is moving backward. The distance from \( t = 8 \) to \( t = 10 \) is:
\[ \text{Distance from } t = 8 \text{ to } t = 10 = s(8) - s(10) = 85.33 - 76.67 = 8.66 \, \text{m \]
Thus, the total distance traveled in 10 seconds is:
\[ D = 85.33 + 8.66 \approx 94 \, \text{m} \]

### Step 8: Find the speed at \( t = 10 \)
Using the velocity function:
\[ v(10) = 16 - 0.25(10^2) = 16 - 25 = -9 \, \text{m/s} \]
The speed (magnitude of velocity) is:
\[ \text{Speed} = 9 \, \text{m/s} \]

### Conclusion
The correct statements are:
- The direction of velocity changes at \( t = 8 \, \text{s} \).
- The distance traveled by the particle in 10 seconds is \( 94 \, \text{m} \).
- The distance traveled in 4 seconds is approximately \( 58.67 \, \text{m} \).
- The speed of the particle at \( t = 10 \, \text{s} \) is \( 9 \, \text{m/s} \).