A particle of unit mass moves so that displacement after $t$ seconds is given by $x = 3 cos (2 t - 4)$ . Find the acceleration and kinetic energy at the end of $2$ seconds.
[ $K . E = \frac{1}{2} m v^{2}, m$ is mass]

Open in App

Solution

Given $x = 3 c o s (2 t - 4), m a s s m = 1$

We have to find the acceleration and kinetic energy at the end of $2$ seconds.

Acceleration is given by $a = \frac{d^{2} x}{d t^{2}}$ and velocity is given by $v = \frac{d x}{d t}$

Consider $x = 3 c o s (2 t - 4)$

Differentiate with respect to $x$ we get

$\frac{d x}{d t} = 3 (- 2 s i n (2 t - 4)) = - 6 s i n (2 t - 4) = v$

Again differentiate with respect to $x$ we get

$\frac{d^{2} x}{d t^{2}} = - 6 (2 c o s (2 t - 4)) = - 12 c o s (2 t - 4) = a$

Thus when $t = 2$ we get

$v = - 6 s i n (4 - 4) = - 6 s i n 0 = 0$ and $a = - 12 c o s (4 - 4) = - 12 c o s 0 = - 12 \times 1 = - 12$

Hence $a = - 12, v = 0$

Kinetic energy is given by $K . E . = \frac{1}{2} m v^{2} = \frac{1}{2} \times 1 \times 0 = 0$

$∴ K . E . = 0, a c c e l e r a t i o n = - 12$ at the end of $2$ seconds.

Suggest Corrections

Similar questions

5

a = 1 m / s^{2}

x

t

x^{2} = 1 + t^{2}

m s^{- 2}

t

2 kg

y (in cm)

t

y = 6 sin (200 t + \frac{π}{6})

1 k g

^{'} y^{'}

^{'} t^{'}

y = 6 s i n 100 t + π / 4

s

t

s = t^{3} - 6 t^{2} + 3 t + 4 m, t

Join BYJU'S Learning Program