Question

A particle of mass 10 kg is moving in a straight line. If its displacement, x with time t is given by x = (t³ – 2t – 10) m, then the force acting on it at the end of 4 seconds is(a) 24 N (b) 240 N(c) 300 N (d) 1200 N

Answer 1

Answer:

The correct answer is option (b) $240 N$.

Explanation:

Concept:

An object's total change in position is referred to as displacement, a vector variable that measures "how far out of place an object is." Displacement is the shift in location over a specific period of time.

Given:

Mass of the particle$=10kg$

It's displacement $x$ with time $t$,

$x = (t^{3}-2t-10) m$

To find:

We have to find the force acting on it at the end of 4 seconds.

Solution:

It is provided that the state of the particle is,

$x = (t^{3}-2t-10) m$

When we differentiating the position with respect to time then we get velocity,

$v=\frac{dx}{dt} \\v=3t^{2} -2$

and by differentiating velocity we get acceleration,

$a=\frac{dv}{dt} \\a=6t$

$a_{t=4} =6*4\\a_{t=4} =24m/s^{2}$

Now for we are finding the force acting at $t=4sec$

$F_{t=4}=ma_{t=4}\\=10*24\\=240N$

Thus, option (b) $240 N$ is the correct answer.

#SPJ3