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The velocity w of a particle in a time t is given by the expression \( \mathrm{W}=109+3 \mathrm{t}^{3} \). Find the instantaneous acceleration after 2 seconds

Ask by Brewer Jimenez. in Nigeria
Mar 02,2025

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Answer

The instantaneous acceleration after 2 seconds is 36 units.

Solution

The instantaneous acceleration is the derivative of the velocity with respect to time. Given the velocity function:   W = 109 + 3t³ Differentiate with respect to time (t):   dW/dt = d(109)/dt + d(3t³)/dt = 0 + 9t² = 9t² At t = 2 seconds, the acceleration becomes:   Acceleration = 9 × (2)² = 9 × 4 = 36 Thus, the instantaneous acceleration after 2 seconds is 36 (units consistent with the given velocity's units, for example, if velocity is in m/s then acceleration is in m/s²).

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To find the instantaneous acceleration, you need to differentiate the velocity function with respect to time. Given the velocity function \( \mathrm{W} = 109 + 3\mathrm{t}^{3} \), the acceleration \( a \) is the derivative \( \frac{dW}{dt} \). Differentiating: \[ a = \frac{dW}{dt} = \frac{d}{dt}(109 + 3\mathrm{t}^{3}) = 0 + 9\mathrm{t}^{2} \] Now, plug in \( t = 2 \) seconds: \[ a(2) = 9(2^{2}) = 9 \times 4 = 36 \, \text{m/s}^{2} \] So, the instantaneous acceleration after 2 seconds is \( 36 \, \text{m/s}^{2} \). Acceleration can often feel mysterious, but it's essentially how quickly your velocity changes. Think of it like speeding up in a car; if you press the gas pedal, your speed increases - that’s acceleration in action!

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