Motion

Speed is the distance travelled per unit time.

Converting distance to SI units: $$s=\frac{9.0\times 1000}{1}=9000\text{ m}$$ Converting time to SI units: $$t=5\times 60=300\text{ s}$$ Equation used — Speed: $$v=\frac{s}{t}$$ Substitution: $$v=\frac{9000}{300}$$ $$v=30{\text{ m s}}^{-1}$$

Speed is the distance travelled per unit time and is a scalar; velocity is speed in a stated direction and is a vector.

The velocity has changed; although the speed remains 4.0 m s⁻¹, the direction has reversed from right to left. Because velocity is speed in a stated direction, a change in direction means a change in velocity.

Average speed = total distance travelled / total time taken.

Step 1: Finding the time for the first section. Converting distance to SI units: $${s}_{1}=\frac{6.0\times 1000}{1}=6000\text{ m}$$ Rearranging for time: $$t=\frac{s}{v}$$ $${t}_{1}=\frac{6000}{20}=300\text{ s}$$ Step 2: Finding the time for the second section. $${s}_{2}=\frac{4.0\times 1000}{1}=4000\text{ m}$$ $${t}_{2}=\frac{4000}{10}=400\text{ s}$$ Step 3: Finding the average speed. $$\text{total distance}=6000+4000=10 000\text{ m}$$ $$\text{total time}=300+400=700\text{ s}$$ $$\text{average speed}=\frac{10 000}{700}$$ $$\text{average speed}=14.3{\text{ m s}}^{-1}\text{ (3 s.f.)}$$

Calculate the gradient of the straight-line section of the graph; divide the change in distance (read from the y-axis) by the change in time (read from the x-axis); the gradient equals the speed.

From 0 s to 5 s the object moves with constant speed; the speed equals 20 ÷ 5 = 4.0 m s⁻¹. From 5 s to 10 s the object is at rest; the distance does not change, so the gradient is zero and the speed is zero.

The area under a speed–time graph represents the distance travelled by the object.

The area under the graph forms a trapezium. $$\text{distance}=\frac{1}{2}\times (a+b)\times h$$ where $a$ and $b$ are the two parallel sides (the two speeds) and $h$ is the time. $$\text{distance}=\frac{1}{2}\times (30+10)\times 8.0$$ $$\text{distance}=\frac{1}{2}\times 40\times 8.0$$ $$\text{distance}=160\text{ m}$$

The object is accelerating; the increasing gradient indicates that the speed is increasing over time.

For the first 6 s the object moves at constant speed of 12 m s⁻¹. For the next 4 s the object decelerates uniformly to rest. Distance during constant speed (rectangle): $${d}_{1}=12\times 6=72\text{ m}$$ Distance during deceleration (triangle): $${d}_{2}=\frac{1}{2}\times 4\times 12=24\text{ m}$$ Total distance: $$d=72+24=96\text{ m}$$

The acceleration of free fall g is approximately 9.8 m/s²; it is approximately constant near the Earth's surface.

In a vacuum there is no air resistance; therefore both objects experience only the gravitational force. The acceleration of free fall g is the same for all objects near the Earth's surface, regardless of mass; so both balls accelerate at the same rate (approximately 9.8 m s⁻²) and reach the ground at the same time.

Acceleration is the change in velocity per unit time.

Converting velocities to SI units: $$u=\frac{36}{3.6}=10{\text{ m s}}^{-1}$$ $$v=\frac{108}{3.6}=30{\text{ m s}}^{-1}$$ Equation used — Acceleration: $$a=\frac{\Delta v}{\Delta t}$$ Given: $$\Delta v=30-10=20{\text{ m s}}^{-1}$$ $$\Delta t=20\text{ s}$$ Substitution: $$a=\frac{20}{20}$$ $$a=1.0{\text{ m s}}^{-2}$$

The gradient of a speed–time graph represents the acceleration of the object.

$$a=\frac{\Delta v}{\Delta t}$$ $$a=\frac{25-5}{10-2}$$ $$a=\frac{20}{8}$$ $$a=2.5{\text{ m s}}^{-2}$$ The acceleration is constant because the section of the graph is a straight line (the gradient does not change).

A deceleration is a negative acceleration; it means the object is slowing down and its velocity is decreasing over time.

Taking the direction of initial motion as positive, the acceleration is −5.0 m s⁻². Rearranging for time: $$\Delta t=\frac{\Delta v}{a}$$ Given: $$\Delta v=0-25=-25{\text{ m s}}^{-1}$$ $$a=-5.0{\text{ m s}}^{-2}$$ Substitution: $$\Delta t=\frac{-25}{-5.0}$$ $$\Delta t=5.0\text{ s}$$