Question

The position vector of a particle changes with time according to the relation ⃗r(t) = 15t² î + (4 - 20t²) jˆ. What is the magnitude of the acceleration at t = 1?
(A) 40 (B) 100
(C) 25 (D) 50

Answer 1

Question

The position vector of a particle changes with time according to the relation ⃗r(t) = 15t² î + (4 - 20t²) jˆ. What is thenmagnitude of the acceleration at t = 1?

(A) 40 (B) 100

(C) 25 (D) 50

Solution

Option (D) is Correct

Given

The position vector of the particle is given as :

$\displaystyle \sf \: r= 15 {t}^{2} \hat{i} + (4 - 20 {t}^{2} ) \hat{j} \\ \\ \leadsto \: \boxed{ \boxed{\sf \: r = 15 {t}^{2} \hat{i} - 20 {t}^{2} \hat{j} + 4 \hat{j}}}$

To finD

Magnitude Of Acceleration

$\rule{300}{2}$

Differentiating x w.r.t to t,we get :

$\sf \: v = \dfrac{dr}{dt} \\ \\ \longmapsto \: \sf \: v = \dfrac{d( {15t}^{2} \hat{i} - 20 \ {t}^{2} \hat{j} + 4 \hat{j}) }{dt} \\ \\ \longmapsto \: \sf \: v = \dfrac{d(15 {t}^{2} \hat{i} )}{dt} - \dfrac{d(20 {t}^{2} \hat{j}) }{dt} + \dfrac{d(4 \hat{j})}{dt} \\ \\ \longmapsto \: \boxed{\boxed{ \sf \: v = (30t \: \hat{i} - 40t \: \hat{j}) \: \: {ms}^{ - 1} }}$

Differentiating v w.r.t t,we get acceleration :

$\sf \: a = \dfrac{dv}{dt} \\ \\ \longmapsto \: \sf \: a = \dfrac{d(30 t \: \hat{i} - 40t \: \hat{j})}{dt} \\ \\ \longmapsto \: \boxed{ \boxed{ \sf \: a = (30 \hat{i} - 40 \hat{j}) \: {ms}^{ - 2} }}$

$\rule{300}{2}$

Now,

$\sf \: |a| = \sqrt{ {(30 \hat{i})}^{2} + ( - 40 \: \hat{j}) {}^{2} } \\ \\ \hookrightarrow \: \sf \sf \: |a| = \sqrt{900 + 1600} \\ \\ \hookrightarrow \: \sf \: |a| = \sqrt{2500} \\ \\ \huge{\hookrightarrow \boxed{ \boxed{ \sf \ |a| = 50 \: { {ms}^{} }^{ - 2} }}}$

$\rule{300}{2}$

$\rule{300}{2}$

Answer 2

Explanation:

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