Greek Lyceum Grade 1 · 60 minutes · 40 marks
Motion in a straight line · Forces and Newton's laws
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A body moves from x=7 m to x=25 m in 6 s. Find average velocity.
Use displacement, not final position: vavg=Δx/Δt.
Δx=25−7=18 m; υμ=18/6=3 m/s
Keep SI units and check the sign.
3 m/s
A body has v₀=14 m/s and acceleration −2 m/s² until it stops. Find stopping time.
Velocity is zero at the stopping instant.
0=14−2t ⇒ t=7 s
Do not extend the equation beyond the stop without a new model.
7 s
On a v–t graph, velocity rises linearly from 0 to 12 m/s in 6 s. Find displacement from area.
The region under the graph is a triangle.
Δx=(6·12)/2=36 m
The units (m/s)·s give m.
36 m
In uniform rectilinear motion, x₀=−7 m and v=6 m/s. Find x at 3 s.
Do not omit the initial position.
x=x₀+υt=−7+3·6=11 m
Displacement is 18 m, different from final position.
11 m
A body of mass 6 kg experiences a horizontal resultant 18 N. Find acceleration.
Newton's second law uses the resultant, not an arbitrary individual force.
a=ΣF/m=18/6=3 m/s²
Keep SI units and check the sign.
3 m/s²
A 6 kg body slides horizontally with μ=0.2 and g=10 m/s². There are no other vertical forces. Find kinetic friction.
Vertical equilibrium gives N=mg; then friction is μN.
N=60 N; T=0.2·60=12 N
Here μN describes kinetic friction, not static friction in general.
12 N
A 7 kg body starts from rest under constant resultant 14 N for 6 s. Find acceleration and final velocity.
Connect dynamics and kinematics: a=ΣF/m, v=v₀+at.
a=2 m/s²; υ=2·6=12 m/s
Constant mass and resultant give constant acceleration.
a=2 m/s²; υ=12 m/s
What resultant is required for acceleration 6 m/s² of a 7 kg body?
Apply ΣF=ma in one inertial reference frame.
ΣF=7·6=42 N
The force points in the required acceleration direction.
42 N
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