Question

a car starting from rest moves with a uniform acceleration of 0.1 ms whole raised to the power -2 for 4 min. find the speed andv the distance travelled

Answer 1

Given :

• Initial velocity ( u ) = 0 m/s

• acceleration ( a ) = 0.1 m/s²

• Time ( t ) = 4 min = 4 x 60 = 240 s

To Find :

• Speed of the car

• Distance travelled by the car

Solution :

$\large \boxed{ \sf \green{ v = u + at}} \\ \\ \implies \sf v = 0 + 0.1 \times 240\\ \\ \boxed{\sf \pink{ v = 24 \: {ms}^{ - 1} }}$

____________________________

$\large \boxed{ \sf \blue{s = ut + \frac{1}{2} a {t}^{2} }} \\ \\ \sf s = 0 \times 240 + \frac{1}{2} \times 0.1 \times {(240)}^{2} \\ \\ \sf s = \frac{1}{2} \times 5760 \\ \\ \large \boxed{ \orange{ \sf s = 2880 \: m}}$

Speed of car = 24 m/s

Distance travelled by car = 2880 m

Answer 2

$\large{\underline{\underline{\mathfrak{Answer :}}}}$

Distance Traveled by car is 2880 m

$\rule{150}{0.1}$

${\underline{\underline{\mathfrak{Step-By-Step-Explanation :}}}}$

$\sf Given \begin{cases} \rm{Accelration \: (a) \: = \: 0.1 \: ms^{-2}} \\ \rm{Time \: (t) \: = \: 4 \: min \: = \: 240s} \\ \rm{Initial \: velocity \: (u) \: = \: 0 \: ms^{-1}} \\ \rm{Distance \: (s) \: = \: ?} \end{cases}$

Use 1 equation of motion :

$\boxed{\boxed{\sf{v \: = \: u \: + \: at}}} \\ \\ \rightarrow {\sf{v \: = \: 0 \: + \: 0.1 \: \times \: 240}} \\ \\ \rightarrow {\sf{v \: = \: 24}} \\ \\ \underline{\sf{\therefore \: Final \: velocity \: is \: 24 \: ms^{-1}}}$

$\rule{200}{2}$

Now, use 3rd equation of motion :

$\boxed{\boxed{\sf{v^2 \: - \: u^2 \: = \: 2as}}} \\ \\ \rightarrow {\sf{(24)^2 \: - \: (0)^2 \: = \: 2(0.1)(s)}} \\ \\ \rightarrow {\sf{576 \: = \: 0.2 \: \times \: s}} \\ \\ \rightarrow {\sf{s \: = \: \dfrac{576}{0.2}}} \\ \\ \rightarrow {\sf{s \: = \: 2880}} \\ \\ \underline{\sf{\therefore \: Distance \: Travelled \: by \: car \: is \: 2880 \: m \: or \: 2.8 \: km}}$