Worked examples · Sheet 1

Uniform rectilinear motions and meeting problems

Theory used: Uniform rectilinear motion Average velocity Meeting problems Reference frame

Exercise 1

On a straight road a cyclist covers the first $12 km$ at the constant speed of $24 km/h$ and the next $12 km$ at the constant speed of $16 km/h$ . Compute the average velocity over the whole route, in $km/h$ and in $m/s$ .

The axis is oriented in the direction of motion; the cyclist never reverses the motion, so the total displacement is $2 d$ , with $d = 12 km$ .

t_{1} = \frac{d}{v _{1}}, t_{2} = \frac{d}{v _{2}}, v_{m} = \frac{2 d}{t _{1} + t _{2}}

v_{m} = \frac{2 d}{\frac{d}{v _{1}} + \frac{d}{v _{2}}} = \frac{2 v _{1} v _{2}}{v _{1} + v _{2}}

v_{m} = \frac{2 \cdot 24 \cdot 16}{24 + 16} km/h = 19.2 km/h = \frac{19.2}{3.6} m/s = 5.33 m/s

In each stretch the motion is uniform, therefore the time taken is the ratio of the length of the stretch to the velocity: $t_{1} = 0.50 h$ and $t_{2} = 0.75 h$ . The average velocity is the total displacement divided by the total time, so it equals $\frac{24 km}{1.25 h}$ . Since the two stretches have the same length, $d$ cancels and the result depends only on the two velocities. Therefore the average velocity is $19.2 km/h$ , that is, $5.33 m/s$ , and it is less than the arithmetic mean $20 km/h$ , because the cyclist spends more time on the slower stretch. As a check, the result lies between the two velocities and is closer to the smaller one.

Exercise 2

On a straight road car $A$ starts from point $P$ at the constant speed of $90 km/h$ ; at the same instant car $B$ starts from point $Q$ , $1200 m$ away from $P$ , and moves toward $A$ at the constant speed of $72 km/h$ . Determine after how much time and at what distance from $P$ the two cars pass each other.

The $x$ -axis is chosen with the origin at $P$ and the positive orientation from $P$ toward $Q$ , and $t = 0$ at the instant of departure; the data are $v_{A} = 25 m/s$ , $v_{B} = 20 m/s$ , $D = 1200 m$ .

x_{A} (t) = v_{A} t, x_{B} (t) = D - v_{B} t

x_{A} (t) = x_{B} (t) ⟺ v_{A} t = D - v_{B} t ⟺ t = \frac{D}{v _{A} + v _{B}}

x = v_{A} t = \frac{v _{A} D}{v _{A} + v _{B}}

t = \frac{1200 m}{45 m/s} = 26.67 s \approx 26.7 s, x = 25 m/s \cdot 26.67 s = 667 m

Equations of motion

x_{A} (t) = 25 t

and

x_{B} (t) = 1200 - 20 t

, in meters: the lines intersect at the instant

t = 26.7 s

at the position

x = 667 m

The two cars must be described in the same reference frame. With the axis oriented from $P$ toward $Q$ , car $B$ moves in the negative orientation, so its velocity is $- v_{B}$ and its equation of motion is $D - v_{B} t$ . The cars pass each other when they occupy the same position, therefore one imposes $x_{A} (t) = x_{B} (t)$ , which is a linear equation in $t$ . The solution $t = \frac{D}{v _{A} + v _{B}}$ shows that the initial distance decreases at the rate $v_{A} + v_{B}$ . The cars pass each other after $26.7 s$ at $667 m$ from $P$ . As a check, $x_{B} = 1200 m - 20 m/s \cdot 26.67 s = 667 m$ , the same value.

Exercise 3

A bus leaves a stop and travels on a straight road at the constant speed of $15 m/s$ . From the same stop, $40 s$ later, a car leaves and travels along the same road at the constant speed of $25 m/s$ . Determine the instant and the position at which the car catches up with the bus.

The $x$ -axis is chosen with the origin at the stop and the orientation of the motion, and $t = 0$ at the departure of the bus; the data are $v_{b} = 15 m/s$ , $v_{a} = 25 m/s$ , $t_{0} = 40 s$ .

x_{b} (t) = v_{b} t, x_{a} (t) = v_{a} (t - t_{0}) for t \geq t_{0}

v_{b} t = v_{a} (t - t_{0}) ⟺ t = \frac{v _{a} t _{0}}{v _{a} - v _{b}}

t = \frac{25 m/s \cdot 40 s}{10 m/s} = 100 s, x = v_{b} t = 15 m/s \cdot 100 s = 1500 m

The car starts with a delay $t_{0}$ , so $t - t_{0}$ appears in its equation of motion, and the law holds only from $t_{0}$ on. The meeting is the solution of $x_{b} (t) = x_{a} (t)$ ; the equation has a positive solution because $v_{a} > v_{b}$ , that is, because the car is faster. The solution $t = 100 s$ is greater than $t_{0}$ , so it is acceptable. Therefore the car catches up with the bus at the instant $t = 100 s$ , $1500 m$ from the stop. As a check, when the car starts the bus is at $600 m$ , and since this distance decreases by $10 m$ every second it becomes zero after $60 s$ , that is, at $t = 40 s + 60 s = 100 s$ .