a bungee jumper leaps off the starting platform at time t=0 and rebounds once during the first five seconds. with velocity measured downward, for t in seconds and 0<t<5, the jumpers velocity is approximated by v(t)= -4t^2 +16t meters/second.
a. how many meters does the jumper travel in the first five seconds?
b. where is the jumper relative to the starting point at the end of five seconds?
c. what does 0-5 v(t) dt represent in terms of the jump?
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- Thread Starter
- 25-04-2016 22:18
- 26-04-2016 01:25
a) v = ds/dt, so s = integral of v dt = 8t^2 - (4/3)t^3 (+c)
However, distance travelled is scalar unlike displacement which is a vector. Therefore, we must take into account which intervals of t give negative v, so we must solve v < 0
v = 16t - 4t^2 = 4t(4-t) = 0 when t = 0 or 4. When t = 2, v = 16 > 0, therefore v < 0 when t < 0 or t > 4. For the interval 0 < t < 5 that we are interested in, this means v is negative for 4 < t < 5
Therefore, distance travelled = (integral btwn 4 and 0 of v dt) - (integral btwn 5 and 4 of v dt) = 128/3 + 28/3 = 156/3 = 52 m
b) The jumpers position is determined by displacement, so s = integral btwn 5 and 0 of v dt = 100/3
Therefore, the jumper is 100/3 m below the starting position.
c) This represents displacement in the downwards direction, relative to the starting position.