Asked by ABAINEH
7. The following table describes the distance covered by a body moving along a straight line to north direction for the first 10 seconds.
All Answers 1
Answered by
GPT-5 mini
AI
I don't see the table — could you paste the distance vs. time values (distance at each second 0–10)? Once you do I can compute and/or plot whatever you need (instantaneous speeds, accelerations, average speed, displacement, velocity–time graph, etc.).
While you paste it, here are the usual steps and formulas I'll use:
- Average speed between t1 and t2:
v_avg = [s(t2) − s(t1)] / (t2 − t1)
- Approximate instantaneous velocity at time ti (for equally spaced data with Δt = 1 s):
central difference: v(ti) ≈ [s(ti+1) − s(ti−1)] / (2·Δt)
forward/backward difference at endpoints:
v(t0) ≈ [s(t1) − s(t0)] / Δt
v(tn) ≈ [s(tn) − s(tn−1)] / Δt
- Approximate acceleration:
a ≈ Δv / Δt, or directly
central second difference: a(ti) ≈ [s(ti+1) − 2·s(ti) + s(ti−1)] / (Δt^2)
- Graphically: plot s vs t — slope = velocity; curvature indicates acceleration. Units: if distances are in metres and time in seconds, velocities in m/s and accelerations in m/s^2.
Paste the table and tell me which results you want (all of the above or specific ones).
While you paste it, here are the usual steps and formulas I'll use:
- Average speed between t1 and t2:
v_avg = [s(t2) − s(t1)] / (t2 − t1)
- Approximate instantaneous velocity at time ti (for equally spaced data with Δt = 1 s):
central difference: v(ti) ≈ [s(ti+1) − s(ti−1)] / (2·Δt)
forward/backward difference at endpoints:
v(t0) ≈ [s(t1) − s(t0)] / Δt
v(tn) ≈ [s(tn) − s(tn−1)] / Δt
- Approximate acceleration:
a ≈ Δv / Δt, or directly
central second difference: a(ti) ≈ [s(ti+1) − 2·s(ti) + s(ti−1)] / (Δt^2)
- Graphically: plot s vs t — slope = velocity; curvature indicates acceleration. Units: if distances are in metres and time in seconds, velocities in m/s and accelerations in m/s^2.
Paste the table and tell me which results you want (all of the above or specific ones).
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