The position vector of a particle changes with time according to the relation r → t = 15 t 2 i ^ + 4 - 20 t 2 j ^ . What is the magnitude of the acceleration at t = 1 ?

r → = 15 t 2 i ^ + 4 - 20 t 2 j ^ v → = 30 t i ^ - 40 t j ^ a → = 30 i ^ - 40 j ^ At t = 1 , a → = 30 i ^ - 40 j ^ a → = 50

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PhysicsKinematics1-DEasy2 minPYQ_2019

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The position vector of a particle changes with time according to the relation $\vec{r} (t) = 15 t^{2} \hat{i} + (4 - 20 t^{2}) \hat{j} .$ What is the magnitude of the acceleration at $t = 1 ?$

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Answer:

Solution:

$\vec{r} = 15 t^{2} \hat{i} + (4 - 20 t^{2}) \hat{j}$
$\vec{v} = 30 t \hat{i} - 40 t \hat{j}$
$\vec{a} = 30 \hat{i} - 40 \hat{j}$
At $t = 1$ ,
$\vec{a} = 30 \hat{i} - 40 \hat{j}$
$|\vec{a}| = 50$

Stream:JEESubject:PhysicsTopic:KinematicsSubtopic:1-D

⏱ 2mℹ️ Source: PYQ_2019

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