(a) Define uniform acceleration. (b) Forces act on a car in motion. List the (i) horizontal forces and their directions; (ii) vertical forces and their dire...
(c) A car starts from rest and accelerate uniformly for 20s to attain a speed of 25 ms\(^{-1}\). It maintains this speed for 30s before decelerating uniformly to rest. The total time for the journey is 60s.
(i) Sketch a velocity-tune graph for the motion.
(ii) Use the graph to determine the (\(\alpha\)) total distance travelled by the car (\(\beta\)) deceleration of the car.
The figure here illustrates force-extension graph for a stretched spiral spring. Determine the work done on the spring.
(a) Definition of uniform acceleration
Uniform acceleration is the constant rate of change of velocity with time; that is, the velocity of the body increases by equal amounts in equal intervals of time.
(b) Forces acting on a car in motion
(i) Horizontal forces and their directions:
Driving force (traction/thrust) – acts forward, in the direction of motion.
Frictional force (including air resistance) – acts backward, opposing the motion.
(ii) Vertical forces and their directions:
Normal reaction from the road – acts upward.
Weight of the car – acts downward.
(c) Motion of the car
The car accelerates from rest to \(25\ \text{m s}^{-1}\) in \(20\ \text{s}\), travels at this steady speed for \(30\ \text{s}\) (i.e. from \(20\ \text{s}\) to \(50\ \text{s}\)), then decelerates uniformly to rest at \(60\ \text{s}\). The deceleration therefore takes \(60 - 50 = 10\ \text{s}\).
(i) Velocity–time graph:
Velocity–time graph: O(0,0)→A(20,25) acceleration, A→B(50,25) constant speed, B→C(60,0) deceleration. Area OABC = total distance = 1125 m.
The graph is the trapezium OABC, where O = (0, 0), A = (20, 25), B = (50, 25) and C = (60, 0).
(ii)(\(\alpha\)) Total distance travelled
The total distance equals the area under the graph (area of the trapezium OABC):
\[ s = \tfrac{1}{2}(AB + OC)\times h = \tfrac{1}{2}\big[(\,50-20\,) + 60\big]\times 25 \]
\[ s = \tfrac{1}{2}(30 + 60)\times 25 = \tfrac{1}{2}\times 90 \times 25 = 1125\ \text{m} \]
Total distance travelled \(= 1125\ \text{m}\).
(ii)(\(\beta\)) Deceleration of the car
The deceleration is the magnitude of the slope of the line BC:
Deceleration of the car \(= 2.5\ \text{m s}^{-2}\).
(d) Work done on the spring
From the force–extension graph, the spring is stretched by a force \(F = 12\ \text{N}\) producing an extension \(e = 0.5\ \text{cm} = \dfrac{0.5}{100} = 0.005\ \text{m}\). The work done equals the area under the force–extension line (a triangle):
Uniform acceleration is the constant rate of change of velocity with time; that is, the velocity of the body increases by equal amounts in equal intervals of time.
(b) Forces acting on a car in motion
(i) Horizontal forces and their directions:
Driving force (traction/thrust) – acts forward, in the direction of motion.
Frictional force (including air resistance) – acts backward, opposing the motion.
(ii) Vertical forces and their directions:
Normal reaction from the road – acts upward.
Weight of the car – acts downward.
(c) Motion of the car
The car accelerates from rest to \(25\ \text{m s}^{-1}\) in \(20\ \text{s}\), travels at this steady speed for \(30\ \text{s}\) (i.e. from \(20\ \text{s}\) to \(50\ \text{s}\)), then decelerates uniformly to rest at \(60\ \text{s}\). The deceleration therefore takes \(60 - 50 = 10\ \text{s}\).
(i) Velocity–time graph:
Velocity–time graph: O(0,0)→A(20,25) acceleration, A→B(50,25) constant speed, B→C(60,0) deceleration. Area OABC = total distance = 1125 m.
The graph is the trapezium OABC, where O = (0, 0), A = (20, 25), B = (50, 25) and C = (60, 0).
(ii)(\(\alpha\)) Total distance travelled
The total distance equals the area under the graph (area of the trapezium OABC):
\[ s = \tfrac{1}{2}(AB + OC)\times h = \tfrac{1}{2}\big[(\,50-20\,) + 60\big]\times 25 \]
\[ s = \tfrac{1}{2}(30 + 60)\times 25 = \tfrac{1}{2}\times 90 \times 25 = 1125\ \text{m} \]
Total distance travelled \(= 1125\ \text{m}\).
(ii)(\(\beta\)) Deceleration of the car
The deceleration is the magnitude of the slope of the line BC:
Deceleration of the car \(= 2.5\ \text{m s}^{-2}\).
(d) Work done on the spring
From the force–extension graph, the spring is stretched by a force \(F = 12\ \text{N}\) producing an extension \(e = 0.5\ \text{cm} = \dfrac{0.5}{100} = 0.005\ \text{m}\). The work done equals the area under the force–extension line (a triangle):
