A vehicle is moving in a straight line. The velocity \( v \mathrm{~m}^{-1} \) at time \( t \) s after the vehicle starts is given by \( v=A\left(t-0.05 t^{2}\right) \) for \( 0 \approx t \approx 15 \) and \( v=\frac{B}{t^{2}} \) for \( t \geqslant 15 \), where \( A \) and \( B \) are constants. The distance travelled by the vehicle between \( t=0 \) and \( t=15 \) is 225 m. i) Find the value of \( A \) and show that \( B=3375 \). \ii) Find an expression in terms of \( t \) for the total distance travelled by the vehicle when \( t \geqslant 15 \). \iii) Find the speed of the vehicle when it has travelled a total distance of 315 m.

To solve this problem, we need to break it down step by step. ### i) Find the value of \( A \) and show that \( B = 3375 \). The distance traveled by the vehicle from \( t = 0 \) to \( t = 15 \) is given by the integral of the velocity function over that time period: \[ d = \int_0^{15} v \, dt = \int_0^{15} A\left(t - 0.05t^2\right) dt \] Calculating the integral: \[ d = A \int_0^{15} \left(t - 0.05t^2\right) dt = A \left[ \frac{t^2}{2} - \frac{0.05t^3}{3} \right]_0^{15} \] Calculating it step by step: \[ = A \left[ \frac{15^2}{2} - \frac{0.05 \cdot 15^3}{3} \right] = A \left[ \frac{225}{2} - \frac{0.05 \cdot 3375}{3} \right] = A \left[ 112.5 - 56.25 \right] = A \left[ 56.25 \right] \] Setting the distance traveled equal to 225 m: \[ A \cdot 56.25 = 225 \implies A = \frac{225}{56.25} = 4 \] Now substituting \( A = 4 \) into the equation for \( v \) when \( t \geq 15 \): At \( t = 15 \): \[ v = 4(15 - 0.05 \cdot 15^2) = 4(15 - 11.25) = 4(3.75) = 15 \, \text{m/s} \] We now use the formula for the velocity from \( t \geq 15 \) to find \( B \): The continuity of velocity at \( t = 15 \) gives us: \[ v = \frac{B}{15^2} = 15 \implies B = 15 \cdot 225 = 3375 \] ### ii) Find an expression in terms of \( t \) for the total distance travelled by the vehicle when \( t \geqslant 15 \). The total distance from \( t = 0 \) to \( t = 15 \) is 225 m. For \( t \geq 15 \), we calculate the distance from \( t = 15 \) to \( t \) as follows: \[ d = \int_{15}^{t} \frac{3375}{u^2} \, du \] Now, integrating: \[ = -\frac{3375}{u} \Big|_{15}^{t} = -\frac{3375}{t} + \frac{3375}{15} \] This means: \[ = -\frac{3375}{t} + 225 \] Thus, the distance for \( t \geq 15 \): \[ \text{Total distance} = 225 + \left(-\frac{3375}{t} + 225\right) = 450 - \frac{3375}{t} \] ### iii) Find the speed of the vehicle when it has travelled a total distance of 315 m. We need to find the value of \( t \) when the total distance is equal to 315 m. Setting the total distance equation to 315 m: \[ 450 - \frac{3375}{t} = 315 \] Rearranging gives us: \[ 450 - 315 = \frac{3375}{t} \] Thus: \[ 135 = \frac{3375}{t} \implies t = \frac{3375}{135} = 25 \, \text{s} \] Now that we have \( t = 25 \), we find the speed at this time using the equation for \( v \): \[ v = \frac{3375}{t^2} = \frac{3375}{25^2} = \frac{3375}{625} = 5.4 \, \text{m/s} \] So, the speed of the vehicle when it has travelled a total distance of 315 m is **5.4 m/s**.