A muon is traveling at 5.00 x 106 m/s when it enters a medium that causes it to decelerate at 1.25 x 1014 m/s2. a) How far does the muon take to stop? b) Graph x versus t and v versus t for the muon.

The initial speed of the muon is $v_0 = 5.00 \times 10^6 \text{ m/s}$.
The muon is slowed, so the acceleration is negative: $a = -1.25 \times 10^{14} \text{ m/s}^2$.
The final speed when the muon stops is $v = 0 \text{ m/s}$.
We need to find the distance $x$. The appropriate kinematic equation is $v^2 = v_0^2 + 2ax$.

Rearrange the equation to solve for $x$:

We use the kinematic equation $v = v_0 + at$.

• Velocity versus time ($v$ vs. $t$): The equation for velocity is $v(t) = v_0 + at$.

This graph is a straight line with a negative slope. It starts at $v = 5.00 \times 10^6 \text{ m/s}$ at $t=0$ and decreases linearly to $v=0 \text{ m/s}$ at $t = 4.00 \times 10^{-8} \text{ s}$.

• Position versus time ($x$ vs. $t$): The equation for position (assuming $x_0=0$) is $x(t) = v_0t + \frac{1}{2}at^2$.

This graph is a parabola opening downwards. It starts at $x=0 \text{ m}$ at $t=0$. The position increases, reaching a maximum value of $0.100 \text{ m}$ at $t = 4.00 \times 10^{-8} \text{ s}$, which is when the muon stops. The slope of the graph (velocity) starts positive and decreases to zero at the maximum displacement.