Explanation:
Given that,
Initial speed of a car, u = 60 km/h = 16.67 m/s
Acceleration, a = 2m/s²
Final speed, v = 120 km/h = 33.33 m/s
We need to find the distance traveled and the time taken to make the distance.
acceleration = rate of change of velocity
let the distance be d.
Hence, the distance traveled and the time taken to make the distance is 208.25 m and 8.33 seconds respectively.
Answer:
distance difference would a) increase
speed difference would f) stay the same
Explanation:
Let t be the time the 2nd skydiver takes to travel, since the first skydiver jumped first, his time would be t + Δt where Δt represent the duration between the the first skydiver and the 2nd one. Remember that as t progress (increases), Δt remain constant.
Their equations of motion for distance and velocities are
Their difference in distance are therefore:
(As
So as time progress t increases, Δs would also increases, their distance becomes wider with time.
Similarly for their velocity difference
Since g and Δt both are constant, Δv would also remain constant, their difference in velocity remain the same.
This of this in this way: only the DIFFERENCE in speed stay the same, their own individual speed increases at same rate (due to same acceleration g). But the first skydiver is already at a faster speed (because he jumped first) when the 2nd one jumps. The 1st one would travel more distance compare to the 2nd one in a unit of time.