Position and Movement Question 1

S(t) = ut + 0.5at^2 where S(t) is a function of t and equal to distance (in metre) travelled after t seconds with given initial speed of u and acceleration of a

for this question,
u = 20ms-1 and a = -1.5ms-2 (deceleration)

After 3 seconds, the distance travelled S(3) = 20 x 3 - 0.5 x 1.5 x 3^2 = 53.25m
After 4 seconds, the distance travelled S(4) = 20 x 4 - 0.5 x 1.5 x 4^2 = 68m

The distance travelled during the 4th second
= distance travelled after 4 seconds - distance travelled after 3 seconds
= S(4) - S(3)
= 68m - 53.25m
= 14.75m
= Ans. C

using s = ut + 1/2at^2

s = 20(4) + 1/2(-1.5) (4)^2

s = 80 - 12

s = 68 m

你係咪抄錯題目 =﹏=|||???

2007-02-26 14:04:57 補充：
係喎! during the 4th s!樓下果個岩呀, 我個錯架。