Physics - Kinematics Question with Solution | TestHub

PhysicsKinematics1-DEasy2 minQB

PhysicsEasynumerical

Velocity of particle depend on x as $v = 4 x^{2} - 4$ where x is positive of the particle then find acceleration at $x = 1$ is.

Answer:

0.00

Solution:

$v = 4 x^{2} - 4$

$\frac{dv}{dx} = 8 x - 4$

at $x = 1$

$\frac{dv}{dx} = 4$

at $x = 1$

$v = 0$

$a = v \frac{dv}{dx} = 0 \times 4 = 0$

Stream:JEESubject:PhysicsTopic:KinematicsSubtopic:1-D

⏱ 2mℹ️ Source: QB

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