SITEMAP   School Physics Notes: Forces & motion 2.4 Advanced motion calculations

UK GCSE level age ~14-16 ~US grades 9-10 Scroll down, take time to study content or follow links

Forces and Motion 2.4 More advanced motion calculations involving uniform acceleration using the formula v2 - u2 = 2as (3rd equation of motion)

Doc Brown's Physics exam study revision notes

2.4 More advanced motion calculations involving uniform acceleration using v2 - u2 = 2as

(this section might not be required for your course)

Constant acceleration is known as linear uniform acceleration.

A good example is an object free falling in a gravitational field e.g. it is the acceleration due to gravity is 9.8 m/s2 at the Earth's surface.

You can use the following formula to do calculations based on a uniform acceleration situation.

An equation for uniform acceleration is ...

(final speed (m/s))2 - (initial speed (m/s))2 = 2 x acceleration (m/s2) x distance (m)

v2 - u2 = 2as, v = final velocity, u = initial velocity, a = acceleration, s = distance travelled or displaced

(This is sometimes referred to as the 3rd equation of motion, 1st equation of motion is v = u + at, and the 2nd equation of motion is s = ut + ½at, time is denoted by t)

Rearrangements to calculate the various object variables in the above equation:

(units above, and remember in algebraic equations,

if you 'change the side you change the sign, or you change all the signs of the expression)

(i) v2 - u2 = 2as,     v2 = 2as + u2,    v = √(2as + u2)

(ii) v2 - u2 = 2as,    -u2 = 2as - v2,    u2 = -2as + v2,    u = √(v2 - 2as)

(iii) v2 - u2 = 2as,   2as = v2 - u2,    a = (v2 - u2) ÷ 2s

(iv) v2 - u2 = 2as,   2as = v2 - u2,    s = (v2 - u2) ÷ 2a

Examples questions based on uniform acceleration

Q3.1 An initially stationary car accelerates uniformly at a rate of 1.50 m/s2.

(a) What is the speed of the car after a distance travelled of 100 m?

(b) If at 100 m the driver presses harder on the accelerator pedal to increase the acceleration to 2.00 m/s, what speed will the car be doing after travelling a total of 300 m from the start?

Worked out ANSWERS to uniform acceleration questions

Q3.2 A jet rocket plane is launched from a converted jet airliner with an acceleration of 10.0 m/s2.

From radar tracking it is found that the rocket plane achieved a speed of 300 m/s after travelling for 1875 m.

At what speed was the launch aircraft travelling?

Worked out ANSWERS to uniform acceleration questions

Q3.3 A car is moving at a speed of 20 m/s when the driver presses on the accelerator pedal and after a distance of 400 m the car is travelling at 30 m/s. Assuming it is uniform, calculate the acceleration of the car.

Worked out ANSWERS to uniform acceleration questions

Q3.4 A rocket is moving vertically upwards and leaves the launch silo with a velocity of 10.0 m/s. Booster rockets are immediately fired to produce an acceleration of 4.00 m/s2.

At what height (s) above the launch pad, will the rocket attain a velocity of 200 m/s (give your final answer in km).

Worked out ANSWERS to uniform acceleration questions

Q3.5 An object is dropped from a height of 10 m above the ground. Ignoring air resistance, if the acceleration due to gravity is 9.8 m/s2,

(a) at what theoretical velocity does the object hit the ground?

(b) If you fall off a ladder at height of just 2 m above the ground (~6 ft) at what speed do you hit the ground at?

Worked out ANSWERS to uniform acceleration questions

Q3.6 A an object is dropped from a certain height above the ground.

If the object hits the ground at 10 m/s, from what height was it dropped?

Worked out ANSWERS to uniform acceleration questions

Q3.7 An object is fired vertically up into the air with an initial velocity of 20.0 m/s.

Ignoring air resistance, and taking the acceleration as -9.80/ms2, calculate the theoretical maximum height the object attains.

Worked out ANSWERS to uniform acceleration questions

Q3.8 A falling object is travelling at 10 m/s when it is 10 m above the ground.

If gravitational acceleration is 9.8 m/s2, at what velocity will it hit the ground?

Worked out ANSWERS to uniform acceleration questions

Keywords, phrases and learning objectives for acceleration and graph work

More advanced motion calculations involving uniform acceleration using the formula v2 - u2 = 2as 3rd equation of motion equation practice exam questions

WHAT NEXT?

BIG website, using the [SEARCH BOX] below, maybe quicker than navigating the many sub-indexes

for UK KS3 science students aged ~12-14, ~US grades 6-8

ChemistryPhysics for UK GCSE level students aged ~14-16, ~US grades 9-10

for pre-university age ~16-18 ~US grades 11-12, K12 Honors

Use your mobile phone in 'landscape' mode?

SITEMAP Website content © Dr Phil Brown 2000+. All copyrights reserved on Doc Brown's physics revision notes, images, quizzes, worksheets etc. Copying of website material is NOT permitted. Exam revision summaries and references to GCSE science course specifications are unofficial.

Using SEARCH some initial results may be ad links you can ignore - look for docbrown

Worked out ANSWERS to the questions based on a uniform acceleration formula

Q3.1 An initially stationary car accelerates uniformly at a rate of 1.50 m/s2.

(a) What is the speed of the car after a distance travelled of 100 m?

v2 - u2 = 2as,     v2 = 2as + u2,    v = √(2as + u2)

Since u = 0 (initially car is stationary) and s = 100 m

v = √(2 x 1.5 x 100) = √300 = 17.3 m/s  (3 s.f.)

(b) If at 100 m the driver presses harder on the accelerator pedal to increase the acceleration to 2.00 m/s, what speed will the car be doing after travelling a total of 300 m from the start?

v2 - u2 = 2as,     v2 = 2as + u2,    v = √(2as + u2)

u = 17.3 m/s (from part (a)), s = 300 - 100 (you must subtract 100 m because that's where new acceleration started

v = √{(2 x  2.0 x 200) + (17.32)} = √(800 + 300) = √1100 =  33.1 m/s  (3 s.f.)

Note the use of ( ) AND { } brackets, you really must take care about what follows the overall square root sign .

Q3.2 A jet rocket plane is launched from a converted jet airliner with an acceleration of 10.0 m/s2.

From radar tracking it is found that the rocket plane achieved a speed of 300 m/s after travelling for 1875 m.

At what speed was the launch aircraft travelling?

v2 - u2 = 2as,    -u2 = 2as - v2,    u2 = -2as + v2,    u = √(v2 - 2as)

Initially both the launch aircraft and the rocket plane are travelling at the same speed u.

v = 300 m/s, a = 10 m/s2, s = 1875 m

u = √(v2 - 2as) = √{(3002) - (2 x 10.0 x 1875)}

u = √(90000 - 37500) = √52500 = 229 m/s  (3 s.f.)

Q3.3 A car is moving at a speed of 20 m/s when the driver presses on the accelerator pedal and after a distance of 400 m the car is travelling at 30 m/s. Assuming it is uniform, calculate the acceleration of the car.

v2 - u2 = 2as,   2as = v2 - u2,    a = (v2 - u2) ÷ 2s

s = 400 m, u = 20 m/s, v = 30 m/s (u and v being the initial and final velocities)

a = (302 - 202) / (2 x 400) = (900 - 400) / 800 = 0.625 = 0.63 m/s2  (2 s.f.)

Q3.4 A rocket is moving vertically upwards and leaves the launch silo with a velocity of 10.0 m/s. Booster rockets are immediately fired to produce an acceleration of 4.00 m/s2.

At what height (s) above the launch pad, will the rocket attain a velocity of 200 m/s (give your final answer in km).

(iv) v2 - u2 = 2as,   2as = v2 - u2,    s = (v2 - u2) ÷ 2a

a 4.00 m/s2, u = 10.0 m/s, v = 200 m/s  (u and v being the initial and final velocities)

s = (2002 - 102) / (2 x 4) = (40000 - 100) / 8 = 39900 / 8 = 4987.5 m = 4990 m or 4.99 km  (3 s.f.)

Note: To fire an object vertically from a 'powerful gun', to completely escape the Earth's gravitational field, requires an 'escape velocity' of 11 km/s at the end of the 'gun barrel'! However, rockets don't function like a gun. They have continuous thrust force from the rocket engines which will sustain a steady upward movement against the force of gravity.

Q3.5 An object is dropped from a height of 10 m above the ground. Ignoring air resistance, if the acceleration due to gravity is 9.8 m/s2,

(a) at what theoretical velocity does the object hit the ground?

v2 - u2 = 2as,     v2 = 2as + u2,    v = √(2as + u2)

u = 0, v = ?, a = 9.8 m/s2, s = 10 m

since u = 0, v = √(2 x 9.8 x 10) = √196 = 14 m/s  (2 s.f.)

Note: Not a good idea to fall when rock climbing, falling from ~31 ft and hitting the ground at 14 m/s. This will do a lot of damage, but no problem if all safety devices in place! BUT, imagine a 100 m sprinter running into a brick wall at an average speed of ~10 m/s!

(b) The general formula for calculating this 'fall' velocity, since u is zero, is:

v = √(2as + u2) = √(2as) = √(2 x 9.8 x height) = √(19.6 x height above ground)

e.g. if you fall off a ladder at height of just 2 m above the ground (~6 ft) you hit the ground at:

v = √(2as) = √(2 x 9.8 x 2) = √39.2 = 6.3 m/s.

Comparing this with the speed of a sprinter, its hardly surprising the serious injuries you can suffer from a relatively short fall.

Q3.6 A an object is dropped from a certain height above the ground.

If the object hits the ground at 10 m/s, from what height was it dropped?

v2 - u2 = 2as, rearranging  s = (v2 - u2) ÷ 2a

s = height, v = 10 m/s, u = 0 m/s, a = 9.8 m/s2 (gravitational acceleration near Earth's surface)

height = s = (102 - 02)  ÷  (2 x 9.8) = 100 ÷  19.6 = 5.1 m

Q3.7 An object is fired vertically up into the air with an initial velocity of 20.0 m/s.

Ignoring air resistance, and taking the acceleration as -9.80/ms2, calculate the theoretical maximum height the object attains.

v2 - u2 = 2as, rearranging  s = (v2 - u2) ÷ 2a

s = (02 - 202)  ÷ (2 x -9.8)

s = -400  ÷ -19.6 = 20.4 m  (to 3 sf)

Q3.8 A falling object is travelling at 10 m/s when it is 10 m above the ground.

If gravitational acceleration is 9.8 m/s2, at what velocity will it hit the ground?

v2 - u2 = 2as,     v2 = 2as + u2,    v = √(2as + u2)

Substituting in the rearranged equation:

v = √{(2 x 9.8 x 10) + 102)

v = √(196 + 100) = √296

final velocity on hitting the ground = 17 m/s (2 sf)

Using SEARCH some initial results may be ad links you can ignore - look for docbrown

 @import url(https://www.google.co.uk/cse/api/branding.css); ENTER specific physics words or courses e.g. topic, module, exam board, formula, concept, equation, 'phrase', homework question! anything of physics interest!  This is a very comprehensive Google generated search of my website

TOP OF PAGE