Kinematics (5) Flashcards
A window cleaner of a block of flats accidentally drops his sandwich, which falls freely to the ground
The speed of the sandwich as it passes a high floor is u ms^-1
After a further 1.2 seconds the sandwich is moving at a speed of 17 ms^-1 past a lower floor
Find the value of u
v = u + at
17 = u + (9.8 * 1.2)
So u = 5.24
A window cleaner of a block of flats accidentally drops his sandwich, which falls freely to the ground
The speed of the sandwich as it passes a high floor is u ms^-1
After a further 1.2 seconds the sandwich is moving at a speed of 17 ms^-1 past a lower floor
The vertical distance between the consecutive floors of the building is h m. It takes the sandwich another 2.1 seconds to fall the remaining 14 floors to the ground. Find h
s = ut + (1/2)at^2
s = (17 * 2.1) + (1/2)(9.8 * 2.1^2)
So s = 57.309
h = s/14 = 4.09m (3 s.f.)
A rocket is projected vertically upwards from a point of 8 m above the ground at a speed of u ms^-1 and travels freely under gravity. The rocket hits the ground at 20 ms^-1
The value of u
v^2 = u^2 + 2as 20^2 = u^2 + (2 * 9.8 * 8) u = ¬(400 - 156.8) = ¬(243.2)
Ignore the negative solution, so u = 15.59487… = 15.6 (3 s.f.)
A rocket is projected vertically upwards from a point of 8 m above the ground at a speed of u ms^-1 and travels freely under gravity. The rocket hits the ground at 20 ms^-1
How long it takes to hit the ground
v = u + at
20 = -15.59487… + 9.8t
So, 9.8t = 35.59487…
Hence t = 3.63212… = 3.63 s (3 s.f.)
A model train sets off from a station at time t = 0 s. It travels in a straight line, then returns to the station.
At time t seconds, the distance, in metres, of the train from the station is s = (1/100)(10t + 9t^2 - t^3) where 0 <= t <= 10.
Sketch the graph of s against t and hence explain the restriction 0 <= t <= 10 s.
s = -(1/100)t(t^2 - 9t - 10) = -(1/100)t(t+1)(t-10)
Draw the sketch
Distance and time can’t be negative so 0 <= t <= 10
A model train sets off from a station at time t = 0 s. It travels in a straight line, then returns to the station.
At time t seconds, the distance, in metres, of the train from the station is s = (1/100)(10t + 9t^2 - t^3) where 0 <= t <= 10.
Find the maximum distance of the train from the station
s is at a maximum when ds/dt = 0, e.g. when v = 0
v = ds/dt = (1/100)(-3t^2 + 18t + 10)
When v = 0, -3t^2 + 18t + 10 = 0
Use the quadratic formula t = 6.51188…
When t = 6.51188…, s = (1/100)(-(6.51188…)^3 + 9(6.51188…)^2 + 10(6.51188…)) = 1.71 m (3 s.f.)
A particle set off from the origin O at t = 0 s and moves in a straight line. At time t seconds, the particle is v ms^-1.
v = 9t - 3t^2 0<=t<=2s v = 24/t^2 t>2s
Find the maximum speed of the particle in the interval 0<=t<=2s
v is at a maximum when dv/dt = 0, e.g. when a = 0
So in the interval 0<=t<=2, a = dv/dt = 9 - 6t
When a = 0: 0 = 9 - 6t, so t = 1.5 s
So: v = (9 * 1.5) - 3(1.5^2)
= 6.75 ms^-1
A particle set off from the origin O at t = 0 s and moves in a straight line. At time t seconds, the particle is v ms^-1.
v = 9t - 3t^2 0<=t<=2s v = 24/t^2 t>2s
t = 2 s t = 6 s
s = ∫ v dt = ((9t^2)/2) - t^3 + C for 0<=t<=2
When t = 0, the particle is at the origin, e.g. s = 0. C = 0
When t = 2, s = (9/2)(2^2) - 2^3 = 10 m
s = ∫ v dt = ∫ 24/t^2 dt = -24/t + C for t>2
When t = 2, s = 10, so 10 = -24/2 + C. C = 22
When t = 6, s = -24/6 + 22 = 18 m
A particle P is moving in a horizontal plane with constant acceleration. After t seconds, P has position vector
[(2t^3 - 7t^2 + 12)i + (3t^2 - 4t^3 - 7)j]m
Where the unit vectors i and j are in the directions of east and north respectively. Find:
An expression for the velocity of P after t seconds
v = s = [(6t^2 - 14t)i + (6t - 12t^2)j] ms^-1
A particle P is moving in a horizontal plane with constant acceleration. After t seconds, P has position vector
[(2t^3 - 7t^2 + 12)i + (3t^2 - 4t^3 - 7)j]m
Where the unit vectors i and j are in the directions of east and north respectively. Find:
The speed of P when t = 0.5, and the direction of motion of P at this time
v = ((6/4) - (14/2))i + ((6/2) - (12/4))j = -5.5i Speed = ¬((-5.5)^2 + 0^2) = 5.5ms^-1
The component of velocity in the direction of north is zero, and the component in the direction of east is negative, so the particle is moving due west
A particle is initially at position vector (i+2j) m, travelling with constant velocity (3i + j) ms^-1. After 8 s it reaches the point A. A second particle has constant velocity (-4i + 2j) and takes 5 s to travel from point A to point B. Find the position vectors of points A and B
A = original position vector + tv = (i + 2j) + 8(3i + j) = (i + 2j) + (24i + 8j) = (25i + 10j) m
B = A + tv = (25i + 10j) + 5(-4i + 2j)
= (25i + 10j) + (-20i + 10j) = (5i + 20j) m
A particle moves in a plane with acceleration a = (12ti - e^(1/3 * t)j) ms^-2. The particle starts at the origin with initial velocity u = (5i - 4j) ms^-1. Find its displacement vector from the origin after 1 second
Integrate a: v = ∫ a dt = (6(t^2)i - 3e^(1/3 * t)j) + C
v = 5i - 4j when t=0 => 5i - 4j = 0 - 3j + C => C = 5i - j
So v = (6t^2 + 5)i - (3e^(1/3 * t) + 1)j ms^-1
Integrate v: s = ∫ v dt = (2t^3 + 5t)i - (9e^(1/3 * t) + t)j + C’
s = 0, when t = 0 => 0 = 0i - 9j + C’ => C’ = 9j
So s = (2t^3 + 5t)i + (9 - 9e^(1/3 * t) - t)j m
When t = 1, s = ( 7i + (8 - 9e^(1/3 * t))j ) m
A stone is thrown from point A on the edge of a cliff, towards a point H, which is on horizontal ground. The point O is on the ground, 11 m vertically below the point of projection. The stone is thrown with speed 15 ms^-1 at an angle a below the horizontal, where tan a = 3/4. The horizontal distance from ) to HJ is 9 m. The stone misses the point H and hits the ground at point B, as shown above.
Find the time taken by the stone to reach the ground
tan a = 3/4 => sin a = 3/5
Vertical motion, taking down as +ve
s = 11 u = u_y = 15 sin a = 9 a = 9.8 t = t
s = ut + (1/2)at^2
11 = 9t + 4.9t^2
Using the quadratic formula: t = 0.839 s (3 s.f.)
A stone is thrown from point A on the edge of a cliff, towards a point H, which is on horizontal ground. The point O is on the ground, 11 m vertically below the point of projection. The stone is thrown with speed 15 ms^-1 at an angle a below the horizontal, where tan a = 3/4. The horizontal distance from ) to HJ is 9 m. The stone misses the point H and hits the ground at point B, as shown above.
Find the horizontal distance the stone misses H by
Horizontal motion, taking right as +ve
s = s u = u_x = 15 cos a = 15 * (4/5) = 12 t = 0.83898... s a = 0
So s = ut => OB = 12 * 0.83898…
OB = 10.06785… m
So it misses H by 10.06785… - 9 = 1.07m (3 s.f.)
A stone is thrown from point A on the edge of a cliff, towards a point H, which is on horizontal ground. The point O is on the ground, 11 m vertically below the point of projection. The stone is thrown with speed 15 ms^-1 at an angle a below the horizontal, where tan a = 3/4. The horizontal distance from ) to HJ is 9 m. The stone misses the point H and hits the ground at point B, as shown above.
Find the speed of the projection which would have ensured that the stone landed at H
Horizontal motion, taking right as +ve
s = 9 u_x = u cos a a = 0 t = t
s = (u_x)t + (1/2)at^2 9 = (u cos a)t => t = 9/(u cos a) (1)
Vertical motion, taking down as +ve s = 11 u_y = u sin a a = 9.8 t = t
s = (u_y)t + (1/2)at^2 => 11 = (u sin a)t + 4.9t^2 (2)
t is the same both horizontally and vertically. So substitute (1) in (2) to eliminate t
11 = 9((u sin a)/(u cos a)) + 4.9(9 / (u cos a))^2
11 = 9 tan a + (4.9 * 81)/(u^2 cos^2 a)
tan a = 3/4 and cos a = 4/5, so substituting and simplifying:
u^2 = 145.919, so u = 12.1 ms^-1 (3 s.f.)
