Sun Earth

Aceleration

Question on Stopping Distances

G and Centripetal Force

Inverse sqaure law

suvat

SUVAT Questions

A cheetah starts from rest and accelerates at 2.0ms^-2 in a straight line for 10s.

 Calculate:

The cheetah’s final velocity

Use of v = u + at → v = 2x10

20m/s                  

The distance the cheetah covers in this 10s

Use of s = ut + ½ at² → s = ½ x 2 x 10²

s = 100m              .

An athlete accelerates out of her blocks at 5.0ms^-2.

How long does it take her to run the first 10m?

Use of s = ut + ½ at² → t² = 2s/a = 20/5 = 4                                                            2 marks

t = 2s

What is her velocity at this point?

Use of v² = u² + 2as → v² = 2x5x10 = 100                                                                2 marks

v = 10m/s

A bicycle’s brakes can produce a deceleration of 2.5ms^-2. Calculate the distance travelled by the bicycle before stopping if it is moving at 10ms^-1 when the brakes are applied?

Use of v² = u² + 2as → 0 = 100 – 2 x 2.5 x s                                                             2 marks

s = 20m

An aircraft is at rest at one end of a runway which is 2.2km long. The aircraft accelerates along the runway with an acceleration of 2.5ms^-2 until it reaches its take-off speed of 75ms^-1.

Calculate:

The time taken to reach take-off speed

Use of v = u + at → 75 = 2.5t                                                                        2 marks

t = 30s  

The distance travelled in this time

Use of s = ut + ½ at² → s = ½ x 2.5 x 30²                                                                            2 marks

s = 1125m

Just as the aircraft reaches take-off speed, a warning light comes on in the cockpit. Reverse thrust can produce a deceleration of 4.0ms^-2.  For reasons he will have to explain at the enquiry, it takes the pilot 2.5s to react, during which time the aircraft continues at its take-off speed. Determine whether the aircraft can stop before it reaches the end of the runway.

Distance travelled before braking = 75 x 2.5 = 187.5m                                    4 marks

Use of v² = u² + 2as → 0 = 75² – 2 x 4 x s                                                                

s = 890 m

Conclusion – yes the runway is long enough since the distance required to stop is smaller than the remaining length of runway (1075m).

The Eagle is landing on the Moon. Neil uses the LEM’s engines to keep its speed of descent constant at 5.0ms^-1 from the time when the craft is 14m above the Moon’s surface until it is 4.0m above the surface. Neil then cuts the engines and lets the Eagle fall freely to the Moon’s surface. The acceleration of free fall near to the Moon’s surface is 1.6ms^-2. Ignore air resistance.   Calculate

The speed of impact

Use of v² = u² + 2as → v² = 5² + 2 x 1.6 x 4                                                              2 marks

v = 6.15 m/s

The time taken to travel the last 4.0m

Use of v = u + at → t = (v – u)/a = (6.15-4)/1.6                                                     2 marks

The time taken for the full 14m descent.

time travelling at constant velocity from s = ut + ½ at² → t = s/u = 10/5 = 2s

Total time = 2.72s

In a test, a car was propelled into gravel trap at 50 ms^-1.  It came to rest in a time of 0.4 s.

Calculate the acceleration? 

-125 ms^-2                                                                                                              1 mark

Calculate the distance travelled before it came to a halt?

Use of s = ut + ½ at² → s = 50x0.4 – ½ x 125 x 0.4²                                                                      2 marks

s = 10m

Robin shoots an arrow at Guy who is riding directly away from him. When he shoots the arrow Guy is 60 m away. When the arrow bounces off Guy’s chainmail, he is 80 m away. If the arrow travels at 70 ms^-1:

Calculate the time taken for the arrow to hit Guy?

Use of s = ut + ½ at²                                                                                                        3 marks

Realises a = 0, s = 80 and u = 70 m/s

t = 80/70 = 1.14s

Calculate Guy’s speed?

v = s/t = 20/1.14                                                                                                                1 mark

v = 17.5 m/s

What assumptions did you make in solving this problem?

The speed of the arrow doesn’t change – air resistance can be ignored.  1 mark

Phileas Fogg’s balloon is gaining height. To avoid crashing into a mountain Passepartout tries to lighten the load by dropping a sandbag from a height of 150m. The hot air balloon is moving upwards with a velocity of 5.0ms^-1. Ignore air resistance.

What is the initial velocity of the sandbag?

5m/s upwards                                                                                                                   1 mark

How long will the bag take to reach the ground?

t = 6.06s

Equations of motion

Speed time graphs

Bullet

velocity - right answers

SPEED AND  VELOCITY

1.     A skier travels 3000m in 93s. Calculate her speed.

S = d ÷ t = 3000m ÷ 93 s = 32.26 m/s

2.     A cyclist travels 1000m in 75s. Calculate his speed.

S = d ÷ t = 1000m ÷ 75 s = 13.33 m/s

3.     A train travels 1200m in 80s. Calculate its speed.

S = d ÷ t = 1200m ÷ 80 s = 15 m/s

4.     A mouse walks 10m in 20s. Calculate its speed

S = d ÷ t = 10m ÷ 20s = 0.5 m/s

5.     Concord travels at 660m/s (twice the speed of sound in air at 0⁰C). How far will it travel in one minute?

d = s x t = 660 m/s x (1 x 60s) = 39600 m = 39.6 km

6.     To leave the Earth’s gravitational field a space vehicle must reach escape speed which is about 18 km/s. How high will the rocket be 34s after lift off? Has any human left the influence of the Earth’s gravitational field?

d = s x t = 18 km/s x 34s = 612 km

No – as the moon orbits the Earth

7.     The term light year is often misused. It is the distance light travels in one year. Given that light travels at 300 000 000m/s (3x 10⁸ m/s) how far is that light year?

1 light year or d = s x t = 3 x 10⁸ m/s x (365.25 days x 24 hr x 60min x 60s) = 9.47 x 10¹⁵ m

If you could travel at the speed of light (not possible) how long would it take you to travel that light year?

For you no time would pass. For those on earth 1 year would pass.

8.     A car travels at 35m/s on a motorway. How long will it take to travel that light year? (In years please)

t = d÷s = 9.47 x 10¹⁵ m ÷ 35 = 2.7 x 10¹⁴ s =  2.7 x 10¹⁴ s ÷ (356.25 days x 24 hr x 60 min x 60 s) = 85.6 million years

9.     A waiter walks at 2m/s. How long will it take him to reach your table 15m away and will your soup be cold when he gets there?

t = d÷s = 15 ÷ 2 m/s = 7.5 s        Yes, it’s Gazpacho

10.  A water ripple travels at 6 cm/s. How long will it take to travel 1m?

t = d÷s = 1 ÷ 6/100 m/s = 16.67s

11.  The average speed of a sprinter is about 11m/s. How long does she take to run the 100m?

t = d÷s = 100 ÷ 11 m/s = 9.09 s (a new world record!)

Compare this speed with a1500m runner who turns in a very respectable time of 3m 50s.

s = d÷ t = 1500m ÷ 230s s = 7.5 m/s

Why is there a difference? Oxygen debt due to build up of lactic acid in muscles

12.  A tennis player’s serve is measured at 50m/s. How long will it take to travel the length of a tennis court (23.8m)

t = d÷s = 23.8m ÷ 50 m/s = 0.48 s

The Steady Earth

I.      The Earth has a diameter of 12756km at the equator. Given it takes 23hrs and 56 minutes to revolve once how fast would you be travelling at the equator? (In m/s and km/h please.)

Distance travelled = circumference of Earth = pd = 3.142 x 12756 = 40 100 km = 4.01 x10⁷ m

Time taken =  23 hrs 56 min = {(23x60) + 56}min x 60s= 86160s

Speed = distance ÷ time = 40100km ÷ 23.93hr = 1676 km/h

Speed = distance ÷ time = 4.01 x 10⁷ m ÷ 86160s = 465.4 m/s

What stops you flying off into space? Gravity

We live some way north of the equator, do we travel as fast? No as we travel less distance in one rotation.

How fast would you be spinning if you were at the North Pole? 0

II.     The average distance from our planet to the sun is 149,503,000 km. The earth orbits the sun once every 365.25 days. How fast are we all going?

Distance travelled = circumference of Earth’s orbit = 2pr = 3.142 x 2 x149 503 000km

= 9.394 x 10⁸ km = 9.394 x 10¹¹m

Time taken =  365.25 days= 365.25 x 24 = 8700 hr = 8700 x 60min x 60s= 3.132 x 10⁷s

Speed = distance ÷ time = 9.394 x 10⁸ km  ÷ 8700 hr = 108 000 km/h

Speed = distance ÷ time = 9.394 x 10¹¹m ÷ 3.132 x 10⁷s = 1.5 x10⁴ m/s = 30 km/s = 30 000 m/s

(Assume the orbit is circular, it's not quite, we are slightly closer to the sun in January than July; if so why is it so cold? The seasons are caused by the tilt of the earth’s axis of rotation to the plane of its orbit)

III.   The sun and the whole solar system are on the move around the centre of our galaxy that is called The Milky Way. We are moving towards the constellation of Hercules at a leisurely 20.1 km/s. At this speed how long would it take to travel around the Earth?

Time = Distance ÷ speed = 40 100 km ÷ 20.1 km/s = 1995 s = 33 min

IV.   This is nothing the whole of our galaxy is on the move as the universe expands. We all move outward towards the constellation of Leo (only from our point of view) at 600km/s. How far have we travelled since you started this worksheet? 360 000km… but then it only took me 10 minutes

Information you may need

Velocity = distance/time distance = speed x time time = distance/velocity

Diameter of a circle = twice its radius (r)              circumference = 2pr       p = 3.142

d-t graphs

weighted test tube

10 SHM Curves

8 Graphs

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