Linear motion under a variable force — practice questions

The force is a function of time, $F(t) = 2t+1$. We use Newton's Second Law, $F=ma$.

$2t+1 = 0.5a$

Since force is a function of time, we use $a = \frac{dv}{dt}$.

$2t+1 = 0.5 \frac{dv}{dt}$

$4t+2 = \frac{dv}{dt}$

To find $v$, we integrate with respect to $t$:

$v = \int (4t+2) \,dt = 2t^2 + 2t + C$

We are given that P is at rest when $t=0$, so $v=0$ when $t=0$. $0 = 2(0)^2 + 2(0) + C \implies C=0$.

So, the velocity at time $t$ is $v = 2t^2 + 2t$.

When $t=4$ s: $v = 2(4)^2 + 2(4) = 2(16) + 8 = 32 + 8 = 40$ m/s.

To find displacement $x$, we use $v = \frac{dx}{dt}$.

$\frac{dx}{dt} = 2t^2 + 2t$

Integrate with respect to $t$:

$x = \int (2t^2 + 2t) \,dt = \frac{2}{3}t^3 + t^2 + D$

We are given that P is at the origin when $t=0$, so $x=0$ when $t=0$. $0 = \frac{2}{3}(0)^3 + (0)^2 + D \implies D=0$.

So, the displacement at time $t$ is $x = \frac{2}{3}t^3 + t^2$.

When $t=4$ s: $x = \frac{2}{3}(4)^3 + (4)^2 = \frac{2}{3}(64) + 16 = \frac{128}{3} + \frac{48}{3} = \frac{176}{3} \approx 58.7$ m (3 s.f.).

Answer: At $t=4$ s, velocity is 40 m/s and displacement is $\frac{176}{3}$ m.

The force is a function of displacement, $F(x) = \frac{18}{x^2}$. The force is directed away from O, so it's in the positive $x$ direction.

Using Newton's Second Law, $F=ma$:

$\frac{18}{x^2} = 2a$

Since force is a function of $x$, we use $a = v\frac{dv}{dx}$.

$\frac{18}{x^2} = 2v\frac{dv}{dx}$

$\frac{9}{x^2} = v\frac{dv}{dx}$

Now, we separate the variables:

$\int \frac{9}{x^2} \,dx = \int v \,dv$

Integrating both sides:

$-\frac{9}{x} + C = \frac{1}{2}v^2$

We are given the initial conditions: at $x=3$, the velocity is $4$ m/s towards O. Since the positive direction is away from O, the velocity is $v = -4$ m/s.

Substitute these values to find C:

$-\frac{9}{3} + C = \frac{1}{2}(-4)^2$

$-3 + C = \frac{1}{2}(16) = 8$

$C = 11$

So the relationship between $v$ and $x$ is: $\frac{1}{2}v^2 = 11 - \frac{9}{x}$.

We need to find the speed when $x=6$. Speed is the magnitude of velocity. Let's find $v^2$.

At $x=6$:

$\frac{1}{2}v^2 = 11 - \frac{9}{6} = 11 - 1.5 = 9.5$

$v^2 = 19$

$v = \pm\sqrt{19}$

The speed is the magnitude of the velocity, which is $|v|$.

Speed = $\sqrt{19} \approx 4.36$ m/s (3 s.f.).

Answer: The speed of the particle at $x=6$ is $\sqrt{19}$ m/s.