The average velocity or displacement of a particle for the first time interval is <u>Δs / Δt = 6 cm/s.</u>
Solution:
As we know that displacement is calculated in centimeters and the unit of time is second.
The average velocity for the first interval [1,2] is given
Δs / Δt = s (t2) - s (t) / t2 - t1
Δs / Δt = 2sin2 π + 3cos 2 π - ( 2sin π + 3cos π ) / 2 - 1
Δs / Δt = 2(0) + 3(1) - 2(0) - 3 (-1) / 1
Δs / Δt = 6 cm/s
Thus the average velocity or displacement of a particle for the first time interval is Δs / Δt = 6 cm/s
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The complete question is:
The displacement of a particle moving back and forth along a line is given by the following equation s(t) = 2sin π t + 3cos π t. Estimate the instantaneous velocity of the particle when t = 1
Work done by the force = Force x displacement. Power = work done/time = F.s/t = F.u.t/t = F.u = 95 x 20 = 1900J. {S=ut because acceleration is zero since car is moving at constant velocity}.
Answer:
Speed: Distance per time, 400 km/h, and a scalar quantity.
Velocity: Displacement per time, 20 m/s south, and a vector quantity.
Explanation:
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Answer:
Plane will 741.6959 m apart after 1.7 hour
Explanation:
We have given time = 1.7 hr
So if we draw the vectors of a 2d graph we see that the difference in angles is = 
Speed of first plane = 730 m/h
So distance traveled by first plane = 730×1.7 = 1241 m
Speed of second plane = 590 m/hr
So distance traveled by second plane = 590×1.7 = 1003 m
We represent these distances as two sides of the triangle, and the distance between the planes as the side opposing the angle 58.6.
Using the law of cosine, 
representing the distance between the planes, we see that:
r = 741.6959 m
