a car travel in straight line with varying velocity given by v=[t-t₀]([ is a modulus)ms^-1 for00..where t is time & t₀ is the time where velocity is zero..the distance travelled is.

To determine the distance traveled by a car moving in a straight line with a varying velocity defined by the equation \( v = |t - t_0| \) m/s, we need to analyze the relationship between velocity, time, and distance. The first step is to understand how the velocity changes over time and then integrate this to find the distance traveled.

Understanding the Velocity Function

The given velocity function \( v = |t - t_0| \) indicates that the velocity depends on the time \( t \) and a specific time \( t_0 \) when the velocity is zero. The modulus function means that the velocity is always non-negative, which implies that the car is always moving forward or stationary at \( t = t_0 \).

Breaking Down the Function

Let's analyze the function in two intervals:

• For \( t < t_0 \): Here, \( v = t_0 - t \) (since \( t - t_0 \) is negative, the modulus makes it positive).
• For \( t \geq t_0 \): In this case, \( v = t - t_0 \) (as \( t - t_0 \) is non-negative).

Calculating Distance Traveled

The distance traveled can be found by integrating the velocity function over the time interval of interest. Let’s assume we want to find the distance traveled from \( t = 0 \) to some time \( t = T \). We will consider two cases based on the value of \( T \) relative to \( t_0 \).

Case 1: \( T < t_0 \)

In this scenario, the velocity function is \( v = t_0 - t \). The distance \( d \) traveled from \( t = 0 \) to \( t = T \) is given by:

d = ∫(t_0 - t) dt from 0 to T

Calculating this integral:

d = [t_0 t - (t^2)/2] from 0 to T = t_0 T - (T^2)/2

Case 2: \( T \geq t_0 \)

Here, we need to split the integral at \( t_0 \) since the velocity function changes. The distance traveled from \( t = 0 \) to \( t = t_0 \) and then from \( t_0 \) to \( T \) is:

d = ∫(t_0 - t) dt from 0 to \( t_0 \) + ∫(t - t_0) dt from \( t_0 \) to \( T \)

Calculating the first integral:

d₁ = [t_0 t - (t^2)/2] from 0 to \( t_0 \) = t_0^2 - (t_0^2)/2 = (t_0^2)/2

Now for the second integral:

d₂ = [ (t^2)/2 - t_0 t] from \( t_0 \) to \( T \) = (T^2)/2 - t_0 T - (t_0^2)/2 + t_0^2 = (T^2)/2 - t_0 T + (t_0^2)/2

Combining both parts gives:

d = (t_0^2)/2 + (T^2)/2 - t_0 T + (t_0^2)/2 = (T^2)/2 - t_0 T + t_0^2

Final Thoughts

In summary, the distance traveled by the car can be expressed as:

• If \( T < t_0 \): \( d = t_0 T - (T^2)/2 \)
• If \( T \geq t_0 \): \( d = (T^2)/2 - t_0 T + t_0^2 \)

This approach allows us to understand how the car's changing velocity affects the total distance traveled over time. By integrating the velocity function, we can capture the nuances of motion in a straightforward mathematical way.