Start Free Mock LoginAlready registered? Login here

Physics - Kinematics Question with Solution | TestHub

PhysicsKinematics1-DMedium2 min

PhysicsMediumsingle choice

Figure shows the plot of velocity versus time on a log-log scale. Assuming straight line motion and the particle to start from origin, the distance covered at the end of $t = 3 s$ is :

Options:

Answer:

Solution:

$lo g v = lo g 2 + (tan 4 5^{\circ}) lo g t [y = mx + c]$

$\Rightarrow lo g v = lo g 2 + lo g t = lo g 2 t \Rightarrow v = 2 t \Rightarrow \frac{d x}{d t} = 2 t \Rightarrow \int_{0}^{x} d x = \int_{0}^{3} 2 t d t \Rightarrow x = (\frac{2 t ^{2}}{2})_{0}^{3} = 9 m$

Stream:JEESubject:PhysicsTopic:KinematicsSubtopic:1-D

⏱ 2m

Doubts & Discussion

Loading discussions...