Question

A car brakes and decelerates uniformly. After t seconds, it has travelled a distance of d metres where d is related to t by the formula, d= 40t-2t^2, t is greater or equal to 10. How far does the car travel in the next second?

Answer 1

Please break it down for me thanks.

Answer 2

i got d=36t+38-2t^2

but the answer is (38-4t) . Please explain why is this so.

Answer 3

ok this equation comes from a physics formula you'll come across:
s = ut + 1/2 a t^2
s = distance
u = initial speed
a = acceleration / deceleration
t = time

so from this we know:
car's initial speed = 40 meters/sec
cars deceleration = -4 meters/sec^2


d= 40t - 2t^2
\[d= 40t-2t^2\] so if t = 10
d = 400 - 200 = 200

in the next second (t = 11) distance =
d= 40t-2t^2
d = 440 - 242
d = 198

Col 1 = Total distance travelled
col 2 = time (seconds)
col 3 = distance travelled since the last second

38 1 38
72 2 34
102 3 30
128 4 26
150 5 22
168 6 18
182 7 14
192 8 10
198 9 6
200 10 2
198 11 -2
192 12 -6
182 13 -10
168 14 -14
150 15 -18
128 16 -22
102 17 -26
72 18 -30
38 19 -34
0 20 -38

Answer 4

so from time = 10 seconds and onwards, it seems like the car is going backwards

Answer 5

as between t = 10 and t = 11, the car travels - 2 m compared to the last second