A particle moves with velocity v = 3t² + 2t. Its acceleration at t = 2 s is:
A. 8 m/s²
B. 16 m/s²
C. 14 m/s²
D. 10 m/s²
Answer: Option C
Solution (By JKSSB Mock Tests)
Acceleration a = dv/dt = d(3t² + 2t)/dt = 6t + 2. At t = 2 s: a = 6(2) + 2 = 14 m/s². Differentiation of velocity gives instantaneous acceleration. Memory tip: a = dv/dt; for polynomial v(t), differentiate term by term. Tests calculus application in kinematics, common in advanced competitive exam questions.
Explanation:
At highest point velocity is zero, but acceleration due to gravity is still g (9.8 m/s²) downward. This causes it to fall back. Acceleration constant throughout free fall.
Explanation:
g is slightly higher at poles (9.83 m/s²) than equator (9.78 m/s²) due to Earth's rotation (centrifugal effect) and equatorial bulge. Weight = mg, so weight increases slightly. Mass constant.
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