Question

The horizontal distances x and vertical height y of a projectile at time t are given by x=at and y=bt²+ct where a,b and c are constants. Then the acceleration of the projectile 1 second after it is fired is​

Answer 1

Horizontal acceleration,

$\displaystyle\longrightarrow\sf {a_x(t)=\dfrac {d^2x}{dt^2}}$

$\displaystyle\longrightarrow\sf {a_x(t)=\dfrac {d^2}{dt^2}(at)}$

$\displaystyle\longrightarrow\sf {a_x(t)=0}$

And vertical acceleration,

$\displaystyle\longrightarrow\sf {a_y(t)=\dfrac {d^2y}{dt^2}}$

$\displaystyle\longrightarrow\sf {a_y(t)=\dfrac {d^2}{dt^2}(bt^2+ct)}$

$\displaystyle\longrightarrow\sf {a_y(t)=2b}$

Then the net acceleration is,

$\displaystyle\longrightarrow\sf {\vec{a(t)}=0\ \hat i+2b\ \hat j}$

$\displaystyle\longrightarrow\sf {\vec{a(t)}=2b\ \hat j}$

And hence the net acceleration acting on the projectile after 1 second of projection is also,

$\displaystyle\longrightarrow\sf {\underline {\underline {\vec{a(1)}=2b\ \hat j}}}$