GCSE level physics exam revision notes: velocity & acceleration

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

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)