Question

l A point moves in xy plane according to equation x = at, y = at(1-bt) where a and b are positive constant and t is the time. the instant at which velocity vector is at π/4 with accleration vector is given by

Answer 1

A point moves in xy plane according to equation, x = at , y = at(1 - bt), where a and b are positive constants.

we know, velocity is the rate of change of displacement with respect to time.

i.e., v = dx/dt

now , x = at

differentiating both sides with respect to t.

dx/dt = d(at)/dt

⇒$v_x$ = a ........(1)

again, y = at(1 - bt) = at - abt²

differentiating both sides with respect to t,

dy/dt = d{at - abt²}/dt

⇒$v_y$ = a - 2abt ......(2)

velocity vector, $\vec{v}=v_x\hat{i}+v_y\hat{j}$

= $a\hat{i}+(a-2abt)\hat{j}$

and for acceleration,

differentiating velocity with respect to t,

so, $\vec{a}={dv_x}{dt}\hat{i}+\frac{dv_y}{dt}\hat{j}$

= (da/dt) i + {da/dt - d(2abt)/dt } j

= 0 - 2ab j

now a/c to question, angle between velocity and acceleration is π/4.

so, cosπ/4 = $\frac{\vec{v}.\vec{a}}{|v|.|a|}$

= {(ai + (a - 2abt )j}.{-2abj}/[√{a² + (a - 2abt)²}} × 2ab]

= -(1 - 2bt)/{√(1 + (1 - 2bt)²}

⇒ 1/√2 = (2bt - 1)/√{1 + (1 - 2bt)²}

⇒1/2 = (2bt - 1)²/{1 + (1 - 2bt)²}

⇒1 + (1 - 2bt)² = 2(2bt - 1)²

⇒1 = (2bt - 1)²

⇒(2bt - 1) = ±1

⇒ 2bt - 1 = 1

⇒ t = 1/b

Answer 2

Answer:

t=1/1*b/1

t=1/b like this all question easy