Worked examples · Sheet 2

Starting and braking with constant acceleration

Theory used: Uniformly accelerated motion Laws of uniformly accelerated motion Relation between velocity and displacement Average acceleration Uniform rectilinear motion Meeting problems

Go to: Exercise 1 Exercise 2 Exercise 3

Exercise 1

A car starts from rest and reaches the speed of $100 km/h$ in $8.0 s$ with constant acceleration. Compute the acceleration and the distance traveled in that time.

The axis is oriented in the direction of motion, with the origin at the starting point and $t = 0$ at the start; the data are $v_{0} = 0$ , $v = 100 km/h = 27.78 m/s$ , $t = 8.0 s$ .

v = v_{0} + a t, x = v_{0} t + \frac{1}{2} a t^{2}

a = \frac{v}{t}, x = \frac{1}{2} \frac{v}{t} t^{2} = \frac{1}{2} v t

a = \frac{27.78 m/s}{8.0 s} = 3.47 m/ s^{2}, x = \frac{1}{2} \cdot 27.78 m/s \cdot 8.0 s = 111 m

The acceleration is constant, so the laws of uniformly accelerated motion hold with $v_{0} = 0$ . The velocity law gives $a = \frac{v}{t}$ , after converting the velocity to $m/s$ by dividing by $3.6$ . Substituting $a$ into the equation of motion, the distance traveled becomes $\frac{1}{2} v t$ , that is, the product of the average velocity $\frac{v}{2}$ and the time. The car accelerates at $3.47 m/ s^{2}$ and travels $111 m$ . The order of magnitude is reasonable, because the acceleration is about one third of $g$ .

Exercise 2

A car is traveling at $90 km/h$ when the driver sees a stationary obstacle on the road. The driver reacts after $0.80 s$ , during which the car continues at constant velocity, and then brakes with a constant deceleration of magnitude $5.0 m/ s^{2}$ . Compute the stopping distance and the stopping time. If the obstacle is at $70 m$ , determine the speed with which the car hits it.

The axis is oriented in the direction of motion, with the origin at the point where the driver sees the obstacle; the data are $v_{0} = 25 m/s$ , $t_{r} = 0.80 s$ , $a = - 5.0 m/ s^{2}$ .

d_{r} = v_{0} t_{r}, 0 = v_{0}^{2} + 2 a d_{f} ⟹ d_{f} = \frac{v _{0}^{2}}{2 ∣ a ∣}, t_{f} = \frac{v _{0}}{∣ a ∣}

d_{r} = 25 m/s \cdot 0.80 s = 20 m, d_{f} = \frac{( 25 m/s ) ^{2}}{2 \cdot 5.0 m/ s ^{2}} = 62.5 m

d = d_{r} + d_{f} = 82.5 m, t = t_{r} + t_{f} = 0.80 s + 5.0 s = 5.8 s

v_{1}^{2} = v_{0}^{2} + 2 a (L - d_{r}) ⟹ v_{1} = v_{0}^{2} - 2 ∣ a ∣ (L - d_{r})

v_{1} = 625 - 2 \cdot 5.0 \cdot 50 m/s = 125 m/s = 11.2 m/s

The motion has two phases. During the reaction time the velocity is constant, so the motion is uniform and the distance traveled is $v_{0} t_{r}$ . During braking the acceleration is constant, and since the time is neither a datum nor needed to find the distance, the relation between velocity and displacement is used with zero final velocity. The braking time is obtained from the velocity law by imposing $v = 0$ . The stopping distance is $82.5 m$ and the stopping time is $5.8 s$ .

The obstacle at $70 m$ is closer than the stopping distance, so the car reaches it while it is still braking, after traveling $L - d_{r} = 50 m$ while braking. The same relation, with the displacement of $50 m$ , gives the speed at impact, $11.2 m/s$ , that is, about $40 km/h$ .

Exercise 3

A car is stopped at a traffic light. When the light turns green it starts with constant acceleration $2.0 m/ s^{2}$ , and at the same instant a truck traveling at the constant speed of $54 km/h$ draws level with it. Determine after how much time and at what distance from the traffic light the car catches up with the truck, the speed of the car at that instant and the maximum distance between the two vehicles before the overtaking.

The origin is chosen at the traffic light, the axis in the direction of motion and $t = 0$ at the start; the data are $a = 2.0 m/ s^{2}$ and $v_{c} = 15 m/s$ .

x_{c} (t) = v_{c} t, x_{a} (t) = \frac{1}{2} a t^{2}

\frac{1}{2} a t^{2} = v_{c} t ⟺ t = 0 or t = \frac{2 v _{c}}{a}

t = \frac{2 \cdot 15 m/s}{2.0 m/ s ^{2}} = 15 s, x = v_{c} t = 225 m, v_{a} = a t = 2 v_{c} = 30 m/s

D (t) = x_{c} (t) - x_{a} (t) = v_{c} t - \frac{1}{2} a t^{2}, D^{'} (t) = v_{c} - a t = 0 ⟺ t = \frac{v _{c}}{a}

D_{m a x} = D (\frac{v _{c}}{a}) = \frac{v _{c}^{2}}{2 a} = \frac{( 15 m/s ) ^{2}}{2 \cdot 2.0 m/ s ^{2}} = 56 m

Equations of motion of the truck,

x_{c} = 15 t

, and of the car,

x_{a} = t^{2}

, in meters: they meet at the instant

t = 15 s

; the dashed segment at the instant

t = 7.5 s

is the maximum distance,

56 m

The truck moves with uniform motion, the car with uniformly accelerated motion starting from rest, and both start from the traffic light at the instant $t = 0$ . The meeting equation has the solution $t = 0$ , which corresponds to the initial drawing level, and the solution $t = 15 s$ , which is the overtaking; at that instant the vehicles are $225 m$ from the traffic light and the car travels at $30 m/s$ , twice the speed of the truck. The result $v_{a} = 2 v_{c}$ is justified by observing that the two vehicles cover the same distance in the same time, so they have the same average velocity, and the average velocity of the car is $\frac{v _{a}}{2}$ .

Before the overtaking the truck is ahead of the car by $D (t)$ . The function $D$ is maximum where its derivative vanishes, that is, when the two velocities are equal. Up to that instant the truck is faster and the distance increases, then the car becomes faster and the distance decreases. Therefore the maximum distance occurs at $t = 7.5 s$ and equals $56 m$ .