The position vector of a particle changes with time according to the relation r(t)= 15 t2 hati + (4 -20 t2) hatj. What is the magnitude of the acceleration (in m/s2) at t = 1?

Question

The position vector of a particle changes with time according to the relation r(t)= 15 t2 hati + (4 -20 t2) hatj. What is the magnitude of the acceleration (in m/s2) at t = 1? (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

vec r = 15t2 hati + (4 - 20 t2) hatj vecv =(d vecr/dt) = 30 t hati - 40 t hatj Acceleration, veca = (d vecv/dt) = 30 hati - 40 hatj ∴ a = √302 + 402 = 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 (in $m / s^{2})$ at $t = 1$ ?

$r = 15 t^{2} \hat{i} + (4 - 20 t^{2}) \hat{j}$
$v = \frac{d r}{d t} = 30 t \hat{i} - 40 t \hat{j}$
Acceleration, $a = \frac{d v}{d t} = 30 \hat{i} - 40 \hat{j}$
$∴ a = 3 0^{2} + 4 0^{2} = 50 m / s^{2}$

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 (in m/s2) at t=1?

r=15t2i^+(4−20t2)j^​ v=dtdr​=30ti^−40tj^​ Acceleration, a=dtdv​=30i^−40j^​ ∴a=302+402​=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 (in $m / s^{2})$ at $t = 1$ ?

$r = 15 t^{2} \hat{i} + (4 - 20 t^{2}) \hat{j}$
$v = \frac{d r}{d t} = 30 t \hat{i} - 40 t \hat{j}$
Acceleration, $a = \frac{d v}{d t} = 30 \hat{i} - 40 \hat{j}$
$∴ a = 3 0^{2} + 4 0^{2} = 50 m / s^{2}$