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Acceleration is defined as the rate of change of velocity. This means a bus can accelerate by changing its speed, direction, or both. However, the most general definition of acceleration involves a change in velocity.

If a body in linear motion changes from point P to Q, the motion is Translational. Translational motion refers to the movement of an object from one location to another in a straight line, without any rotation involved.

the car is moving with uniform acceleration.

Mean acceleration = \(\frac{\Delta V}{\Delta t}\) Point 1: (2s, 5m/s) Point 2: (4s,15m/s) Mean acceleration = \(\frac{ 15 - 5 }{4 - 2}\) = 5.0ms\(^2\)

Distance = Speed \timesTime Distance = 335 x 5 = 1675m

The magnitude of the acceleration of a moving object is equal to the rate of change of its velocity over time. Gradient of a velocity-time graph refers to the slope of the graph that plots velocity (y-axis) against time (x-axis).

The total area under the entire segment from P to R (the area of the triangle plus the area of the rectangle) represents the total distance traveled by the cyclist during the motion from P to R.

A body moves with uniform velocity when there is no acceleration. According to Newton's first law of motion, an object will remain in its state of rest or uniform motion unless acted upon by a net external force.Therefore, if the resultant force acting on the body is equal to zero, it will maintain a constant velocity, which means it can be moving at a constant speed in a straight line.

Total mass of the two boxes = 27.3 + 6.2 = 33.5kg F = ma a = \(\frac{\text{F}}{\text{m}}\) = \(\frac{150}{33.5}\) = 4.4776... ≈ 4.5 ms\(^{-2}\)

(a)i c = 50.0cm (ii) M\(_R\) = 90.0 g (vii) d = 0.8m (ix) S/N h(cm) t(s) T= \(\frac{t}{20}\)s T\(^2\)s\(^2\) 1 40.0 26.12 1.306 1.7056 2 50.0 27.80 1.390 1.9321 3 60.0 30.68 1.534 2.3532 4 70.0 34.60 1.730 2.9929 5 80.0 35.80 1.790 3.2041 (x) SEE THE GRAPH ABOVE (xi) Slope, s = \(\frac{\DeltaT^2}{\Delta h}\) = \(\frac{3.0 - 1.75}{75 - 46}\) = \(\frac{1.25}{29}\) = 0.431s\(^2\)cm\(^{-1}\) (xii) K = \(\frac{\text{S}}{\text{Q}}\) where, Q = \(\frac{2}{25d^2}\) = \(\frac{2}{25 \times (0.8)^2}\) = 0.125 K = \(\frac{\text{S}}{\text{Q}}\) = \(\frac{0.431}{0.125}\) = 0.3448 b(i) A couple, as related to oscillatory motion, refers to a pair of forces that are equal in magnitude but opposite in direction, and act on opposite sides of a pivot or fulcrum, causing rotation or torsion. b(ii) I. Wrenches: When using a wrench to tighten or loosen a bolt, the force applied creates a couple that produces rotational motion. II. Steering Wheels: Turning the steering wheel of a car applies a couple that rotates the front wheels, allowing for directional change. III. Door Handles: When pushing or pulling on one side of a door handle, a couple is created that rotates the door about its hinges. IV. Bicycle Pedals: When pedaling a bicycle, the force applied to the pedals generates a couple that turns the crankshaft, propelling the bike forward.

R = \(\frac{u^2 sin 2\theta}{g}\) From the above formula, it is evident that range depends on acceleration due to gravity. For real, the range of a projectile is affected by the acceleration due to gravity; it influences how far the projectile travels horizontally.

Given: time of flight T = 4s, g = 10ms\(^{-2}\), H\(_{max}\) \(\theta\) = ? T = \(\frac{2u sin\theta}{g}\) 4 = \(\frac{2 \times u sin \theta}{10}\) 40 = 2u sin\(\theta\) sin\(\theta\) = \(\frac{20}{u}\) But, H\(_{max}\) = \(\frac{u^2 sin^2 \theta}{2g}\) H\(_{max}\) = \(\frac{u^2 \times (\frac{20}{u})^2}{2 \times 10}\) H\(_{max}\) = \(\frac{20 \times 20}{20}\) = 20 metres.(u\(^2\)s cancel out) Thus H\(_{max}\) attained = 20metres.

The man's weight = T - ma mass = 60kg, a = 3ms\(^2\) The man's weight = m(g - a) = 60(10 - 3) = 60 x 7 = 420N

For the law of conservation of momentum to hold, the system must be closed (no mass enters or leaves) and isolated (no external forces acting on it).

Given: mass of boy = 40 kg = m\(_b\), mass of skateboard = ? = m\(_k\), speed of skateboard v\(_x\) = - 20 m/s, speed of boy v\(_y\) = 1.5 m/s Both the skateboard and the boy were initially at rest. The total momentum before and after the jump must be equal. 0 = m\(_b\) x v\(_y\) + m\(_k\) x v\(_x\) 0 = 40 x 1.5 + m\(_k\) x (-20) 0 = 60 - 20m\(_k\) 20m\(_k\) = 60 m\(_k\) = \(\frac{60}{20}\) = 3 kg.