Question

If force $F=(500-100t)$ , then impulse as a function of time will be:
A) $500t-50{{t}^{2}}$
B) $50t-10$
C) $50-{{t}^{2}}$
D) $100{{t}^{2}}$

Answer 1

A force is any interaction that, when unopposed, will change the motion of an object. A force can change the state of rest or the state of motion of an object. We generally conceive force in the form of a push or a pull. Impulse is created when a force acts on a body for a small period.

Formula Used:
$F=\dfrac{dP}{dt}$ and $\int{dP}=\int{Fdt}$

Complete step by step solution:
In classical physics, the impulse is the integral of a force $F$ , over a time interval $t$ .
Mathematically we can say that $F=\dfrac{dP}{dt}$ where $dP$ is the change in impulse over an infinitesimal time interval $dt$ .
Taking the denominator to the left-hand side, we can say that $dP=Fdt$
Now, integrating both sides, we can say that $\int{dP}=\int{Fdt}$
Substituting the values, we can express the relation as $\int{dP}=\int{(500-100t)dt}$
Solving this integration, we get $P=500t-100\left( \dfrac{{{t}^{2}}}{2} \right)+c$
Simplifying the above result, we can say that $P=500t-50{{t}^{2}}+c$ , where $c$ is the Integration constant
Initially, that is, at the time $t=0$ force $F=500$
Since force is constant, the impulse must be $0$
Satisfying the above condition into the impulse expression obtained, we deduce that the value of Integration constant $c=0$.

Hence, option (A) is the correct answer.

Note: Alternatively, the impulse can be calculated by finding the change in the momentum of the object. Since force is equal to mass times acceleration and acceleration can be expressed as the rate of change of velocity with time. Through cross multiplication of the denominator, we get that force times the change in time is equal to mass times the change in velocity, that is, $F\Delta t=m\Delta v$ .