car B travels at a constant speed of 90 km/h. while car A accelerates uniformly 0 to 150 km/h , reaches top speed after 5x10^ -3 hours, then travels at a constant speed speed. what is the acceleration rate of car A and when will car A reach car B?
2007-09-09 14:21:09 · 1 answers · asked by Anonymous in Science & Mathematics ➔ Physics
A accelerates at a = 150/0.005 (km/hr^2).
Note m/s^2 = km/hr^2 * m/km * (hr/s)^2 = km/hr^2 * 0.00007716,
m/s = km/hr * m/km * hr/s = km/hr * 0.27777777...
define t1=intercept time
Average speed of A over acceleration phase = 150/2 = 75. Distance of A over acceleration phase = 75*0.005 = 0.375. Average speeds of A and B are equal (=90) at the overtaking point. Since 75 < 90, overtaking occurs after A has reached vmax.
s = 90t1 = 0.375 + 150(t1-0.005)
= 0.375 + 150t1 - 0.75; t1 = 0.375 / 60 = 0.00625 km
2007-09-10 06:16:36 · answer #1 · answered by kirchwey 7 · 0⤊ 0⤋