Answer:
40 s
Explanation:
After 10 seconds, the first skater would have a 8m/s * 10s = 80 m head start
Let t be the number of seconds after the second skater starts will the second skater overtake the first skater
The distance traveled by the first skater after t seconds is
Similarly the distance traveled by the 2nd skater after t seconds is
Since the 2nd skater catches up to the 1st one after 80 m behind, the distance traveled by the 2nd one must be 80m greater than the distance of the 1st skater
We can substitute 
Answer:
a) X = 17.64 m
b) X = 17.64 + 4∆t^2 + 16.8∆t
c) Velocity = lim(∆t→0)〖∆X/∆t〗 = 16.8 m/s
Explanation:
a) The position at t = 2.10s is:
X = 4t^2
X = 4(2.10)^2
X = 17.64 m
b) The position at t = 2.10 + ∆t s will be:
X = 4(2.10 + ∆t)^2
X = 17.64 + 4∆t^2 + 16.8∆t m
c) ∆X is the difference between position at t = 2.10s and t = 2.10 + ∆t so,
∆X= 4∆t^2 + 16.8∆t
Divide by ∆t on both sides:
∆X/∆t = 4∆t + 16.8
Taking the limit as ∆t approaches to zero we get:
Velocity =lim(∆t→0)〖∆X/∆t〗 = 4(0) + 16.8
Velocity = 16.8 m/s
Answer:
100/10 = 10 , 10 × 10 = 100÷20 = 5
I'm pretty sure its wrong
