Answer:
Correct answer: a = 15.46 m/s²
Explanation:
The formula for accelerated movement with the given data is:
V² - V₀² = 2 · a · d where the initial velocity V₀ and the final V
Since the initial velocity V₀ is zero, the formula is:
V² = 2 · a · d => a = V² / (2 · d) = 27.8² / (2 · 25) = 772.84 / 50 = 15.46 m/s²
a = 15.46 m/s²
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Answer:
Explanation:
Given the equation modelled by the height of the train given as:
s(t) = 18t²-2t³ for for 0 ≤ t ≤ 9
a) Velocity is the rate of change of displacement.
Velocity = dS(t)/dt
V = dS(t)/dt = 36t - 6t² miles
Velocity at t = 3hrs is determiner by substituting t = 3 into the velocity function.
V = 36(3) -6(3)²
V= 108 - 72
Velocity = 36mi/hr
b) for Velocity at time = 7hrs
V(7) = 36(7) - 6(7)²
V(7) = 252 - 294
V(7) = -42mi/hr
The velocity at t = 7hrs is -42mi/hr
c) Acceleration is the rate of change of velocity.
a(t) = dV(t)/dt
Given v(t) = 36t - 6t²
a(t) = 36 - 12t
Acceleration at t=1 is given as:
a(1) = 36 -12(1)
a(1) = 24mi/hr²
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Answer:
3. 1.8A
Explanation:
Given the following data;
Quantity of charge, Q = 650C
Time = 6 minutes to seconds = 6 * 60 = 360 seconds.
To find the current l;
Quantity of charge = current * time
Substituting into the equation, we have;
650 = current * 360
Current = 650/360
Current = 1.8 Amperes