JEE MAIN - Physics (2020 - 4th September Morning Slot - No. 3)
Starting from the origin at time t = 0, with initial velocity 5$$\widehat j$$ ms-1 , a particle moves in the x-y plane
with a constant acceleration of $$\left( {10\widehat i + 4\widehat j} \right)$$ ms-2. At time t, its coordinates are (20 m, y0
m). The
values of t and y0 are, respectively:
5s and 25 m
2s and 18 m
2s and 24 m
4s and 52 m
Vysvětlení
$$y = {u_y}t + {1 \over 2}{a_y}{t^2}$$
$$y = 5t + {1 \over 2}(4){t^2}$$
$$y = 5t + 2{t^2}$$
and $$x = 0(t) + {1 \over 2}(10)({t^2}) = 20$$
$$t = 2s$$
$$ \Rightarrow y = 10 + 8 = 18m$$
$$y = 5t + {1 \over 2}(4){t^2}$$
$$y = 5t + 2{t^2}$$
and $$x = 0(t) + {1 \over 2}(10)({t^2}) = 20$$
$$t = 2s$$
$$ \Rightarrow y = 10 + 8 = 18m$$