Answer:
3 m/s
Explanation:
<u>The velocity of a position-time graph is the slope of the line.</u> Slope is rise over run, or rise divided by run. The rise (how many units the line goes up) is 3 units and the run is 1 unit. 3/1 is 3, so the velocity is 3 m/s.
Hence ,From the Guide there are other parameters which with this equation will give the exact time the athlete's walk back

From the question we are told
If the average velocity during the athlete's walk back to the starting line in Guided Example 2.5 is – 1.50 m/s,
Generally the equation Time spent is mathematically given as
T=\frac{d}{v}
Therefore

Hence ,From the Guide there are other parameters which with this equation will give the exact time the athlete's walk back

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D. 980, this is the best answer because 35 x 7 is 980 :)
Answer:
(A) Q = 321.1C (B) I = 42.8A
Explanation:
(a)Given I = 55A−(0.65A/s2)t²
I = dQ/dt
dQ = I×dt
To get an expression for Q we integrate with respect to t.
So Q = ∫I×dt =∫[55−(0.65)t²]dt
Q = [55t – 0.65/3×t³]
Q between t=0 and t= 7.5s
Q = [55×(7.5 – 0) – 0.65/3(7.5³– 0³)]
Q = 321.1C
(b) For a constant current I in the same time interval
I = Q/t = 321.1/7.5 = 42.8A.