Question

A train accelerates from 36km/h to 54km/h in 10 sec. Find acceleration and the distance travelled by the car.​

Answer 1

Answer:

Initial speed u = 36 km/h = 36 x 5/18 m/s = 10 m/s

Final speed v = 54 km/h = 54 x 5/18 m/s = 15 m/s

Time t = 10 s

Acceleration a = (v - u)/t = (15 - 10)/10 = 5/10 = 0.5 m/s2.

Distance travelled  S = ut + ½ at2

                                    = 10 x 10 + ½ x 0.5 x 10 x 10     m

                                    = 100 + 25 = 125 m

Answer 2

Answer:

Answer :

• The acceleration of car is 0.5 m/s.
• The distance traveled by car is 125 m.

$\begin{gathered}\end{gathered}$

Given :

• Initial velocity = 36km/h
• Final velocity = 54km/h
• Time = 10 second

$\begin{gathered}\end{gathered}$

To Find :

• Acceleration
• Distance travelled

$\begin{gathered}\end{gathered}$

Using Formulas :

${\longrightarrow{\small{\underline{\boxed{\sf{a = \dfrac{v - u}{t}}}}}}}$

${\longrightarrow{\small{\underline{\boxed{\sf{S = ut + \dfrac{1}{2}{at}^{2} }}}}}}$

• v = final velocity
• u = initial velocity
• a = acceleration 
• t = time taken.
• s = distance

$\begin{gathered}\end{gathered}$

Solution :

✒ Coverting the Initial velocity and final velocity into m/s :

Initial velocity

$\rightarrow{\sf{Initial \:  velocity = 36 k/m}}$

$\rightarrow{\sf{Initial \:  velocity = 36 \times \dfrac{5}{18}}}$

$\rightarrow{\sf{Initial \:  velocity =  \cancel{36} \times \dfrac{5}{\cancel{18}}}}$

$\rightarrow{\sf{Initial \:  velocity = 2 \times 5}}$

$\rightarrow{\sf{Initial \:  velocity = 10m/s}}$

${\rightarrow{\underline{\boxed{\sf{\red{Initial \:  velocity = 10m/s}}}}}}$

Final velocity :

$\rightarrow{\sf{Final  \: velocity = 54 k/m}}$

$\rightarrow{\sf{Final  \: velocity = 54 \times  \dfrac{5}{18}}}$

$\rightarrow{\sf{Final \: velocity =  \cancel{54} \times \dfrac{5}{\cancel{18}}}}$

$\rightarrow{\sf{Final \: velocity = 3 \times 5}}$

$\rightarrow{\sf{Final \: velocity = 15m/s}}$

${\rightarrow{\underline{\boxed{\sf{\red{Final \: velocity = 15m/s}}}}}}$

Hence, the Initial velocity and final velocity is 10m/s and 15m/s.

$\rule{300}{1.5}$

✒ Finding the acceleration :

${\rightarrow{\sf{Acceleration = \dfrac{v - u}{t}}}}$

${\rightarrow{\sf{Acceleration = \dfrac{15 - 10}{10}}}}$

${\rightarrow{\sf{Acceleration = \dfrac{5}{10}}}}$

${\rightarrow{\sf{Acceleration = \cancel{\dfrac{5}{10}}}}}$

${\rightarrow{\sf{Acceleration = {\dfrac{1}{2}}}}}$

${\rightarrow{\sf{Acceleration = 0.5 m/s}}}$

$\bigstar \: \underline{\boxed{\sf{\red{Acceleration = 0.5 m/s}}}}$

Hence, the acceleration of car is 0.5 m/s.

$\rule{300}{1.5}$

✒ Now, finding the distance traveled by the car :

${\rightarrow{\sf{Distance = ut + \dfrac{1}{2}{at}^{2}}}}$

${\rightarrow{\sf{Distance = \bigg(10 \times 10 \bigg)+ \dfrac{1}{2}{ \bigg(0.5 \times {(10)}^{2} \bigg)}}}}$

${\rightarrow{\sf{Distance =100+ \dfrac{1}{2}{ \bigg(0.5 \times {(10 \times 10)} \bigg)}}}}$

${\rightarrow{\sf{Distance =100+ \dfrac{1}{2}{ \bigg(0.5 \times 100 \bigg)}}}}$

${\rightarrow{\sf{Distance =100+ \dfrac{1}{2} \times 50}}}$

${\rightarrow{\sf{Distance =100+ \dfrac{1}{\cancel{2}} \times \cancel{50}}}}$

${\rightarrow{\sf{Distance =100 + 25}}}$

${\rightarrow{\sf{Distance =125 \: m}}}$

${\bigstar \: {\underline{\boxed{\sf{\red{Distance =125 \: m}}}}}}$

Hence, the distance traveled by car is 125 m.

$\begin{gathered}\end{gathered}$

Learn More :

Velocity :

${\longrightarrow{\small{\underline{\boxed{\bf{v = u  +  at}}}}}}$

Displacement with positive accelerations :

${\longrightarrow{\small{\underline{\boxed{\bf{s= ut  +  \dfrac{1}{2}{at}^{2}}}}}}}$

Displacement with negative acceleration :

${\longrightarrow{\small{\underline{\boxed{\bf{s= ut - \dfrac{1}{2}{at}^{2}}}}}}}$

Displacement knowing initial and final speeds :

${\longrightarrow{\small{\underline{\boxed{\bf{s= \dfrac{1}{2}\big(u + v \big)t}}}}}}$

Velocity squared :

${\longrightarrow{\small{\underline{\boxed{\bf{{v}^{2} =  {u}^{2} + 2as}}}}}}$

${\color{gray}{\rule{220pt}{2.5pt}}}$