Question

A particle is moving with velocity V=4t^3+3t^2-1m/s. The displacement of particle in time
t = 1s to t= 2 s will
a) 13 m
b) 9 m
c) 17m
d)21 m​

Answer 1

Explanation:

option d 21 meter is the answer

Answer 2

Answer:

The correct option is d) 21 m.

Explanation:

As per the question,

We need to find the displacement of the particle.

As we know,

• Displacement can be obtained from the function of velocity by integrating it in the terms of t, which is time.
• So we can write

$Displacement=\int\limits^2_1 {dv} \, dt \\\\\ Displacement= \int\limits^2_1 {4t^{3}+3t^{2}-1} \, dt \\\\Displacement=|t^{4} +t^{3} -t|^2_1$

• Therefore the equation of displacement in this case is now obtained and by putting the value of the from t=1s to t=2s we will get:

$=|t^{4} +t^{3} -t|^2_1\\=(16+8-2) - (1)\\=22-1\\=21$

Hence,

The correct option is d) 21 m.

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