The position vector of a particle changes with time according to the relation vecr (t) = 15t2 hati + ( 4 -20 t2) hatj. What is the magnitude of the acceleration at t = 1 ?

Question

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

Tardigrade Admin · Accepted Answer

vecr = 15t2 hati +(4-20 t2) hatj vecv = (d vecr/dt) =30 t hati + (-40t) hatj veca = (d vecv/dt) =30 hati - 40 hatj | veca| = 50 m/s2

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

40

100

25

50

$r = 15 t^{2} \hat{i} + (4 - 20 t^{2}) \hat{j}$
$v = \frac{d r}{d t} = 30 t \hat{i} + (- 40 t) \hat{j}$
$a = \frac{d v}{d t} = 30 \hat{i} - 40 \hat{j}$
$∣ a ∣ = 50 m / s^{2}$

Q. The position vector of a particle changes with time according to the relation r(t)=15t2i^+(4−20t2)j^​. What is the magnitude of the acceleration at t=1 ?

40

100

25

50

r=15t2i^+(4−20t2)j^​ v=dtdr​=30ti^+(−40t)j^​ a=dtdv​=30i^−40j^​ ∣a∣=50m/s2

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

$r = 15 t^{2} \hat{i} + (4 - 20 t^{2}) \hat{j}$
$v = \frac{d r}{d t} = 30 t \hat{i} + (- 40 t) \hat{j}$
$a = \frac{d v}{d t} = 30 \hat{i} - 40 \hat{j}$
$∣ a ∣ = 50 m / s^{2}$