A ball, dropped vertically, falls d metres in t seconds. d is proportional to the square of t. The ball drops 45 metres in the first 3 seconds. How far does the ball drop in the next 7 seconds?
First we form equations from the information given in the question. The first equation we can form is that d is directly proportional to the square of t. This means that d = kt2 ,where k is a constant of proportionality. Using the given relationship of d = 45 when t = 3, we can work out the constant k. 45 = k3245 = 9kk = 5The relationship between d and t is now d = 5t2 We are asked for the next 7 seconds, so we cannot simply substitute t = 7 into this equations. If we did this then we would be working out the first 7 seconds.Instead we do 3 + 7 = 10 to work out the total time.We will work out the distance dropped in 10 seconds and then subtract the first 3 seconds worth of distance. When t = 10, d = 5102 = 500Then we subtract the first 3 seconds worth of distance, 500 - 45 = 455455 is the distance travelled by the ball in the next 7 seconds.
