Very difficult question on speed/distance/time can someone explain it to me!? maths..?
The average speed for the first 6 seconds was 30 mph. 30 mph is 1/2 mile/minute, so in 6 seconds he traveled 1/20 of a mile. So he traveled 39/20 mile at 1 mile/minute and 1/20 of a mile at 1/2 mile/minute. (39/20 mile) / (1 mile/minute) = 39/20 minutes (1/20 mile) / (1/2 mile/minute) = 1/10 minute (39/20) + (2/20) = 41/20 minutes (41/20 minutes) x (60 seconds/minute) = 123 seconds
Can you please tell me how to solve this tough "Speed-distance-time" question?
here is the question: A boy goes from his house to his school daily by a motorbike. He drives the motorbike at three different speeds ‘a’, ‘2.5a’ and ‘4a’ km/hr for three different durations during the journey. He definitely knows the time intervals for which he should drive his motorbike at each of these speeds so as to reach the school on time. On one particular day he starts 20 minutes late from his house and hence the extra time for which he drives his motorbike at ‘2.5a’ km/hr is 6 minutes. Find the extra time for which he drives at ‘4a’ km/hr to reach the school on time. options are:- a)11/3 minutes b)15/4 minutes c)13/3 minutes d)cannot be determined i will post the answer tomorrow..but i don't know how to solve it. can i apply ratio and proportion technique here if yes then how?
Hard algebra problems, involving speed, distance and time? Please help!!?
1) speed = M/T km per min = (M/T)(1/60) = M / (60T) km per sec Time = distance / speed = N / (M / 60T) = N * (60T / M = (60 NT) / M 2) Total Distance = M + N Total Time = (M/V) + (N/U) Average speed = (total Distance) / (total Time) = (M + N) / (M/V + N/U) 3) Right now: Boy = x Sister = x + 3 Three years ago: Boy = x - 3 Sister = (x + 3) - 3 = x (x - 3) = (2/3)(x) 3(x - 3) = 2x 3x - 9 = 2x x = 9 Boy is now 9 and sister is now 12
Distance/Time/Speed word problems?
1 - For this question you can utilise the equation v = u + at .... where v is final velocity, u is initial velocity, t is time between u & v, and a is acceleration (in this case due to gravity) .... ==> 22 = 0 + 9.8t ==> t = 22/9.8 = 2.244897959 seconds (choose acuracy as desired) 2 - since speed = distance / time (s = d/t) we can work out the distance each person ran as follows .... Pam -- 6 = d/(25/60) ==> d = 6(25/60) ==> d = 2.5 miles Jill -- 5.5 = d/(25/60) ==> d = 5.5(25/60) ==> d = 55/24 miles Nancy -- 5 = d/(45/60) ==> d = 5(45/60) ==> d = 3.75 miles Average Speed for the 3 = total distance / total time ==> Average Speed = (2.5 + 55/24 + 3.75) / (95/60) ==> Average Speed = (205/24) / (95/60) ==> Average Speed = (205/24)(60/95) ==> Average Speed = 5 & 15/38 (5.3947 to 4dp)
Can you solve this hard physics question?
When fired horizontally, a bullet will fall the same amount as if you dropped it, so you can use the normal equations of motion. 1. Find the time as a function of height dropped and g (gravity) h = ½gt² t = √2h/g t = √(2 * 0.049 / 9.8) t = 0.1 s 2. Calculate the distance: x = vt x = 655 * 0.1 x = 65.5 m