Define (i) velocity and (ii) friction - Leaving Cert Physics - Question a - 2017
Question a
Define (i) velocity and (ii) friction.
A car started from rest and accelerated at 0.4 m s² to reach a top speed of 28 m s⁻¹; it maintained this speed for 200 second... show full transcript
Worked Solution & Example Answer:Define (i) velocity and (ii) friction - Leaving Cert Physics - Question a - 2017
Step 1
(i) Define velocity
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Answer
Velocity is defined as the rate of change of displacement with respect to time in a specified direction. It is a vector quantity, which means it has both magnitude and direction.
Step 2
(ii) Define friction
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Friction is the force that opposes the motion of two surfaces that are in contact. It acts in the opposite direction to the applied force, and its magnitude depends on the nature of the surfaces and the normal force pressing them together.
Step 3
(iii) Draw a diagram indicating the main forces acting on the car when it was accelerating.
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In the diagram, the following forces should be indicated:
Weight (W) acting downwards
Reaction force (R) from the ground acting upwards
Driving force from the engine (F_engine) acting forwards
Frictional force (F_friction) acting backwards. The driving force should exceed the friction for the car to accelerate.
Step 4
(iv) Calculate how long it took the car to reach its top speed.
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To find the time taken to reach the top speed, we can use the formula for acceleration:
\frac{28 \, ext{m/s} - 0 \, ext{m/s}}{t} = 0.4 \, ext{m/s}^2\
\Rightarrow t = \frac{28 \, ext{m/s}}{0.4 \, ext{m/s}^2} = 70 \, ext{s}$$
Thus, it took the car 70 seconds to reach its top speed.
Step 5
(v) Sketch the velocity-time graph for the journey.
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The velocity-time graph would show:
A linear increase in velocity from 0 to 28 m/s over 70 seconds (acceleration).
A horizontal line from 70 seconds to 270 seconds (constant speed).
A linear decrease from 28 m/s to 0 over the final 30 seconds (deceleration). The graph should reflect these three phases clearly.
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