Movement And Position

the object moves at constant speed for the first 10 s the gradient is constant so speed does not change the object is stationary from 10 s to 20 s because the distance remains unchanged

Finding the average speed in the first section $${v}_{1}=\frac{150}{5.0}=30\text{ m/s}$$ Finding the average speed in the second section $${v}_{2}=\frac{150}{15}=10\text{ m/s}$$ The object moves three times faster in the first section because the gradient is three times steeper.

distance travelled per unit time

Equation used $$v=\frac{d}{t}$$ Given $$d=2.7\text{ km}$$ $$t=4.5\text{ min}$$ Working Converting distance from km to m: $$d=2.7\times 1000$$ $$d=2700\text{ m}$$ Converting time from minutes to seconds: $$t=4.5\times 60$$ $$t=270\text{ s}$$ Substituting: $$v=\frac{2700}{270}$$ $$v=10{\text{ m s}}^{-1}$$ Answer $$v=10{\text{ m s}}^{-1}$$

speed = distance ÷ time, so speed is inversely proportional to time if time doubles and distance stays the same, the speed becomes half its original value

measure and record the time taken to travel fixed distance intervals plot distance on the y-axis against time on the x-axis if the graph is a straight line, the car moves at constant speed if the line curves and becomes steeper, the car is accelerating

Finding the mean time $${t}_{\text{mean}}=\frac{1.8+2.1+1.9}{3}$$ $${t}_{\text{mean}}=\frac{5.8}{3}$$ $${t}_{\text{mean}}=1.933\text{ s}$$ Finding the average speed Equation used $$v=\frac{d}{t}$$ Given $$d=2.0\text{ m}$$ $$t=1.933\text{ s}$$ Substituting: $$v=\frac{2.0}{1.933}$$ $$v=1.03...\approx 1.0{\text{ m s}}^{-1}\text{ (2 s.f.)}$$ Repeating measurements and calculating the mean reduces the effect of random errors; such as variations in reaction time when using a stopwatch.

the change in velocity per unit time it is a vector quantity / has direction

Taking the direction of travel as positive: Equation used $$a=\frac{v-u}{t}$$ Given $$u=90\text{ km/h}$$ $$v=0{\text{ m s}}^{-1}$$ $$t=50\text{ s}$$ Working Converting initial velocity from km/h to m/s: $$u=\frac{90}{3.6}$$ $$u=25{\text{ m s}}^{-1}$$ Substituting: $$a=\frac{0-25}{50}$$ $$a=-0.50{\text{ m s}}^{-2}$$ The negative sign indicates deceleration (opposite to the direction of travel). Answer $$a=-0.50{\text{ m s}}^{-2}\text{ (deceleration of }0.50{\text{ m s}}^{-2}\text{)}$$

acceleration is inversely proportional to time when the change in velocity is constant so doubling the time halves the acceleration

acceleration

the distance travelled

Finding the distance during acceleration (trapezium area) $${d}_{1}=\frac{1}{2}\times (5.0+25)\times 8.0$$ $${d}_{1}=\frac{1}{2}\times 30\times 8.0$$ $${d}_{1}=120\text{ m}$$ Finding the distance during constant velocity (rectangle area) $${d}_{2}=25\times 12$$ $${d}_{2}=300\text{ m}$$ Finding the total distance $${d}_{\text{total}}=120+300$$ $${d}_{\text{total}}=420\text{ m}$$ Answer $${d}_{\text{total}}=420\text{ m}$$

Taking the initial direction of travel as positive: Finding the deceleration $$a=\frac{v-u}{t}$$ $$a=\frac{0-18}{6.0}$$ $$a=-3.0{\text{ m s}}^{-2}$$ The deceleration is 3.0 m/s². Finding the distance (triangle area) $$d=\frac{1}{2}\times 18\times 6.0$$ $$d=54\text{ m}$$ Answer $$a=-3.0{\text{ m s}}^{-2}\text{ (deceleration of }3.0{\text{ m s}}^{-2}\text{)}$$ $$d=54\text{ m}$$

Equation used $${v}^{2}={u}^{2}+2as$$ Rearranging for acceleration: $$a=\frac{{v}^{2}-{u}^{2}}{2s}$$ Given $$u=36\text{ km/h}$$ $$v=90\text{ km/h}$$ $$s=175\text{ m}$$ Working Converting initial speed from km/h to m/s: $$u=\frac{36}{3.6}$$ $$u=10{\text{ m s}}^{-1}$$ Converting final speed from km/h to m/s: $$v=\frac{90}{3.6}$$ $$v=25{\text{ m s}}^{-1}$$ Substituting: $$a=\frac{{25}^{2}-{10}^{2}}{2\times 175}$$ $$a=\frac{625-100}{350}$$ $$a=\frac{525}{350}$$ $$a=1.5{\text{ m s}}^{-2}$$ Answer $$a=1.5{\text{ m s}}^{-2}$$

Finding the final speed Equation used $${v}^{2}={u}^{2}+2as$$ Given $$u=0{\text{ m s}}^{-1}$$ $$a=9.8{\text{ m s}}^{-2}$$ $$s=4.5\text{ m}$$ Substituting: $${v}^{2}={0}^{2}+2\times 9.8\times 4.5$$ $${v}^{2}=88.2$$ $$v=\sqrt{88.2}$$ $$v=9.39...\approx 9.4{\text{ m s}}^{-1}\text{ (2 s.f.)}$$ Finding the time The output of the previous step (v = 9.39 m/s) feeds into the acceleration equation. Equation used $$a=\frac{v-u}{t}$$ Rearranging for time: $$t=\frac{v-u}{a}$$ Substituting (using the unrounded value from the previous step): $$t=\frac{9.39-0}{9.8}$$ $$t=0.958...\approx 0.96\text{ s (2 s.f.)}$$ Answer $$v\approx 9.4{\text{ m s}}^{-1}$$ $$t\approx 0.96\text{ s}$$