A driver spotted a police car and he braked from a speed of 100km/h to a speed of 80km/h during a displacement of 88m, at a constant acceleration (a) What is that acceleration. (b) How much time is required for the given decrease in speed ​

Let's first transform (I am not an English speaker and I cannot find a better word) the units to SI-units so we can work with them: V1 = 100 km/h = 27.8 m/s and V2 = 88 km/h = 24.4 m/s. That means that ∆V = V2 - V1 = 24.4 m/s - 27.8 m/s = 3.3m/s (always positive). Now is the average speed between those speeds the same as (27.8m/s+24.4m/s)/2 = 26.1 m/s. v=x/t and that means that t=x/v so t=88m /26.1m/s = 3.4s . Now is a = ∆v/t = 3.3m/s / 3.4s = 0.99m/s².

Answer: a = − 1.579 [tex]\frac{m}{s^{2} }[/tex]b. t = 3.52sWe have velocity in km/hr and displacement in m. Since we have to make a unit conversion, might as well do it so it will yield the answer in the normal (SI) units for acceleration.u = 100 x [tex]\frac{5}{18}[/tex] = 27.78 [tex]\frac{m}{s}[/tex]v = 80 x [tex]\frac{5}{18}[/tex] = 22.22 [tex]\frac{m}{s}[/tex]Using the formula of motion [tex]v^{2} - u^{2}[/tex] = 2*a*d[tex](22.22)^{2}[/tex] - [tex](27.78)^{2} \\[/tex] = 2 * a * 88a = -1.579 [tex]\frac{m}{s^{2} }[/tex] = accelerationSince the acceleration was constant, we can use the average velocity, and the distance to determine the time.time = [tex]\frac{displacement}{speed}[/tex]t = [tex]\frac{88}{speed}[/tex] speed = [tex]\frac{1}{2}[/tex] * (u+v) = [tex]\frac{1}{2}[/tex] * (22.22 + 27.78) = 25[tex]\frac{m}{s}[/tex]t = [tex]\frac{88}{25}[/tex] = 3.52 sTo learn more about speed and acceleration,https://brainly.com/question/25332191