Question

ܬ A 0.150 kg Particle move along x axis according to x(t) -13+2t +4t*2+3t*3 with x in meters time in seconds. What is the net force on particle particle at t= 2.60 sec?​

Answer 1

Answer:

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Explanation:

F=ma

to get a, we need to differentiate x(t) twice

$\frac{dx}{dt} = v = 9 {t}^{2} + 8t + 2 \\ \frac{dv}{dt} = a = 18t + 8 \\ at \: t = 2.6s \: \: \: a = 46.8 + 8 = 54.8 \: m {s}^{ - 2}$

now F= 0.15×54.8= 8.22N