Question

A particle is moving with a velocity of v =(3i^+4tj^)v=(3i^+4tj^) m/s. Find the ratio of tangential acceleration to that of normal acceleration at t = 1 sec.

Answer 1

Answer:

solved

Explanation:

v = dx/dt i^ + dy/dt j^ = 3i^+4tj^

dx/dt   = 3

dy/dt = 4t

d²x/dt ²  = 0

d²y/dt² = 4

Answer 2

The ratio of tangential acceleration to that of normal acceleration at t=1 s=4: 3

Explanation:

Velocity of particle , v$=(3\hat{i}+4t\hat{j})m/s$

t=1 s

$a=\frac{dv}{dt}$

$a=\frac{d(3i+4tj}{dt}=4\hat{j}$

Substitute t=1

Let $v=v(1)=3\hat{i}+4\hat{j}$

$\mid v\mid=\sqrt{x^2+y^2}$

Where x= Coefficient of unit vector i

y=Coefficient of unit vector j

$\mid v\mid=\sqrt{3^2+4^2}=\sqrt{25}=5$

$a\cdot v=4\hat{j}\cdot (3\hat{i}+4\hat{j})=16$

Because $\hat{i}\cdot \hat{i}=\hat{j}\cdot\hat{j}=\hat{k}\cdot\hat{k}=1,i\cdot j=0$

$a\times v =4\hat{j}\times (3\hat{i}+4\hat{j})=-12\hat{k}$

$i\times i=j\times j=k\times k=0,i\times j=k,j\times k=i,k\times i=j,j\times i=-k,i\times k=-j,k\times j=-i$

$\mid a\times v\mid=\sqrt{(-12)^2}=12$

Tangential component of acceleration, $a_T=\frac{a\cdot v}{\mid v\mid}$

Tangential acceleration, at t=1 s $a_T=\frac{16}{5}m/s^2$

Normal component of acceleration , $a_N=\frac{\mid a\times v\mid}{\mid v\mid}$

Normal component of acceleration at t=1 , $a_N=\frac{12}{5} m/s^2$

Ratio of tangential acceleration to the normal acceleration at t=1 s

$\frac{a_T}{a_N}=\frac{\frac{16}{5}}{\frac{12}{5}}=\frac{4}{3}$

Hence, the ratio of tangential acceleration to that of normal acceleration at t=1 s=4: 3

#Learns more:

A particle moves in a circle of radius 1.0 cm at a speed given by v=2.0t where v is in cm/s and and t in seconds.

(a) Find the radial acceleration of the particle at t=1 s. (b) Find the tangential acceleration at t=1 s (c) Find the magnitude of the acceleration at t = 1 s.

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