To solve the problem step by step, we will break it down into manageable parts.
### Step 1: Understand the motion of the car
The car starts from rest with an initial acceleration of \(6 \, \text{m/s}^2\) which decreases linearly to \(0 \, \text{m/s}^2\) over \(10\) seconds. After \(10\) seconds, the car moves at a constant speed.
### Step 2: Determine the acceleration function
Since the acceleration decreases linearly from \(6 \, \text{m/s}^2\) to \(0 \, \text{m/s}^2\) in \(10\) seconds, we can express acceleration \(a(t)\) as:
\[
a(t) = 6 - \frac{6}{10}t = 6 - 0.6t
\]
where \(t\) is in seconds.
### Step 3: Find the velocity function
To find the velocity, we need to integrate the acceleration function:
\[
v(t) = \int a(t) \, dt = \int (6 - 0.6t) \, dt = 6t - 0.3t^2 + C
\]
Since the car starts from rest, \(v(0) = 0\), so \(C = 0\). Therefore, the velocity function is:
\[
v(t) = 6t - 0.3t^2
\]
### Step 4: Calculate the velocity at \(t = 10\) seconds
Now, we substitute \(t = 10\) seconds into the velocity function:
\[
v(10) = 6(10) - 0.3(10^2) = 60 - 30 = 30 \, \text{m/s}
\]
Thus, after \(10\) seconds, the car reaches a constant speed of \(30 \, \text{m/s}\).
### Step 5: Calculate the distance covered in the first \(10\) seconds
Next, we need to find the distance covered during the first \(10\) seconds. We integrate the velocity function:
\[
s(t) = \int v(t) \, dt = \int (6t - 0.3t^2) \, dt = 3t^2 - 0.1t^3 + C
\]
Again, since the car starts from rest, \(s(0) = 0\), so \(C = 0\). Thus, the distance function is:
\[
s(t) = 3t^2 - 0.1t^3
\]
Now, we calculate the distance at \(t = 10\) seconds:
\[
s(10) = 3(10^2) - 0.1(10^3) = 300 - 100 = 200 \, \text{m}
\]
### Step 6: Determine the remaining distance
The total distance to be covered is \(400 \, \text{m}\). The distance covered in the first \(10\) seconds is \(200 \, \text{m}\), so the remaining distance is:
\[
400 \, \text{m} - 200 \, \text{m} = 200 \, \text{m}
\]
### Step 7: Calculate the time to cover the remaining distance
After \(10\) seconds, the car travels at a constant speed of \(30 \, \text{m/s}\). The time \(t_r\) required to cover the remaining \(200 \, \text{m}\) is given by:
\[
t_r = \frac{\text{distance}}{\text{speed}} = \frac{200 \, \text{m}}{30 \, \text{m/s}} = \frac{20}{3} \, \text{s} \approx 6.67 \, \text{s}
\]
### Step 8: Calculate the total time
The total time \(T\) taken to travel \(400 \, \text{m}\) is:
\[
T = 10 \, \text{s} + t_r = 10 \, \text{s} + \frac{20}{3} \, \text{s} = 10 + 6.67 \approx 16.67 \, \text{s}
\]
### Final Answer
The total time required for the car to travel \(400 \, \text{m}\) from the start is approximately \(16.67 \, \text{s}\).
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