A particle moves towards east with velocity $5\ m/s$ . After $10\ seconds$ its direction changes towards north with same velocity. The average acceleration of the particle is
Zero
$\frac{1}{{\sqrt 2 }}\ m/s^2\ N-W$
$\frac{1}{{\sqrt 2 }}\ m/s^2\ N-E$
$\frac{1}{{\sqrt 2 }}\ m/s^2\ S-W$
The position of a particle is given by
$r=3.0 t \hat{i}+2.0 t^{2} \hat{j}+5.0 \hat{k}$
where $t$ is in seconds and the coefficients have the proper units for $r$ to be in metres.
$(a)$ Find $v (t)$ and $a (t)$ of the particle.
$(b)$ Find the magnitude and direction of $v (t)$ at $t=1.0 s$
The height $y$ and the distance $x$ along the horizontal plane of a projectile on a certain planet (with no surrounding atmosphere) are given by $y = (8t - 5{t^2})$ meter and $x = 6t\, meter$, where $t$ is in second. the angle with the horizontal at which the projectile was projected is
A particle is moving along a curve. Then
A person walks $25.0^{\circ}$ north of east for $3.18 \,km$. How far would she have to walk due north and then due east to arrive at the same location?
The position of a projectile launched from the origin at $t=0$ is given by $\vec{r}=(40 \hat{i}+50 \hat{j}) m$ at $t=$ $2 s$. If the projectile was launched at an angle $\theta$ from the horizontal, then $\theta$ is (take $g =10\,ms ^{-2}$ )