Question

A cat is moving at 18m/s when it accelerates for 2 seconds. What is it’s acceleration?

Answer 1

Explanation:

Here, u = 0 m/s, v = 18 m/s, t = 2 s, a =?

By first equation of motion:

$v = u + at \\ 18 = 0 + 2 \times 2 \\ \implies \: 2a = 18 \\ \implies \: a = \frac{18}{2} \\ \implies \: a = 9 \: m \: {s}^{ - 2}$

Answer 2

Explanation:

$\begin{gathered} \tt \: Force \: acting \: on \: mass \: m_1 \\ F = m_1a \\ \implies \: m_1 = \frac{F}{a} = \frac{2}{5} \: kg \\ \tt Force \: acting \: on \: mass \: m_2 \\ F = m_2a \\ \implies \: m_2 = \frac{F}{a} = \frac{2}{7} \: kg \ \\ \tt \: \ now \: when \: masses \: are \: tied \: together \: then \: \\ \rm \: total \: mass = m_1 + m_2 \\ = \frac{2}{5} + \frac{2}{7} \\ = \frac{14 + 10}{35} = \frac{24}{35} \: kg \\ \tt let \: the \: acceleration \: produced \: be \: a \\ \bf \: hence \\ a = \frac{F}{m_1 + m_2} = \frac{2}{ \frac{14}{35} } = \frac{2 \times 35}{14} \\ = \frac{2 \times 7 \times 5}{7 \times 2} = 5 \: m/ {s}^{2} \\ thus \: a= 5 \: m/ {s}^{2} \\ \end{gathered}$