10 questions · Form 5 Additional Mathematics Bab 8: Kinematics of Linear Motion
When an object moves with uniform velocity, what is its acceleration?
Prefer reading to quizzing? All 10 questions are listed below with the answer and explanation under each one.
1. When an object moves with uniform velocity, what is its acceleration?
Answer: A
Uniform velocity means velocity is constant, so its derivative dv/dt = a = 0.
2. An object passes through fixed point O when t = 0. What is its displacement at t = 0?
Answer: A
By definition, the fixed reference point O has displacement s = 0.
3. How is acceleration obtained from a displacement function s(t)?
Answer: A
Velocity v = ds/dt and acceleration a = dv/dt = d²s/dt².
4. What is the condition for a particle to be instantaneously at rest?
Answer: A
A particle is at rest when its velocity v is equal to 0.
5. The acceleration of an object is given by a = 6t. If initial velocity is 0, find displacement s as a function of t (assuming s = 0 at t = 0).
Answer: A
v = ∫ 6t dt = 3t² + c (since v(0)=0, c=0). s = ∫ 3t² dt = t³ + d (since s(0)=0, d=0). Thus, s = t³.
6. If displacement s = -15 m, what does this indicate about the position of the particle?
Answer: A
Negative displacement means position is to the left or negative side of the reference origin O.
7. Given acceleration a = 6 ms⁻². If initial velocity is 2 ms⁻¹, find the velocity after 4 seconds.
Answer: A
Since a is constant, v = u + at = 2 + (6)(4) = 2 + 24 = 26 ms⁻¹.
8. What does a negative acceleration (a < 0) represent?
Answer: A
Negative acceleration indicates that velocity is decreasing with respect to time, which is deceleration.
9. What is the difference between displacement and total distance?
Answer: A
Displacement measures net change in position (vector), whereas total distance sums up all physical movement along the path (scalar).
10. A particle moves such that its velocity is v = 4 - 2t. What is the total distance traveled from t = 0 to t = 3 s?
Answer: A
Rest point v = 0 => 4 - 2t = 0 => t = 2 s. s = ∫ (4 - 2t) dt = 4t - t². At t = 0: s = 0. At t = 2: s = 4(2) - (2)² = 4 m. At t = 3: s = 4(3) - (3)² = 3 m. Distance = |4 - 0| + |3 - 4| = 4 + 1 = 5 m.