Question

Please answer it as, fast as u can! A train accelerates from 36 km/h to 54 km/h in 10 sec. (i) Acceleration (ii) The distance travelled by car.

Answer 1

$\huge \tt \underline \orange {Answer}$

$\sf\pink{Initial \: Velocity (u) = 36 km/hr = 10 m/s}$

$\sf\pink{Final \: Velocity (v) = 54 km/hr = 15m/s}$

$\sf\pink{Time = 10 sec}$

Acceleration = $\huge\boxed{\tt{ \frac{v - u}{t} }}$

$\implies \frac{15 - 10}{10} \\ \\ \implies \frac{5}{10} \\ \\ \implies0.5$

Thus, Acceleration = 0.5 m/s²

Distance Travelled (S) = $\tt using \rightarrow \boxed{\tt{ {v}^{2} = {u}^{2} + 2as }} \\ \tt \implies {15}^{2} = {10}^{2} + 2 \times \frac{1}{2} \times s \\ \tt \implies 225 = 100 + s \\ \tt \implies s = 225 - 100 \\ \tt \implies s = 125 m$

Therefore,

$\bf\green{Acceleration = 0.5 m/s^{2}}$

$\bf\green{Distance \: Travelled = 125 m}$

$\huge \tt \underline \purple {Additional \: Information}$

Acceleration is a measure of change in the Velocity of an object per unit time.

SI unit → m/s²

• What is the diffrence between acceleration and momentum

$\longrightarrow$ Acceleration is the diffrence between final and initial velocity.

And Momentum, is the product of Mass and velocity.

Answer 2

${\underline{\sf{Question}}}$

A train accelerates from 36 km/h to 54 km/h in 10 sec. (i) Acceleration (ii) The distance travelled by ca

${\purple{\boxed{\large{\bold{Formula's}}}}}$

Kinematic equations for uniformly accelerated motion .

$\bf\:v=u+at$

$\bf\:s=ut+\frac{1}{2}at{}^{2}$

$\bf\:v{}^{2}=u{}^{2}+2as$

and $\bf\:s_{nth}=u+\frac{a}{2}(2n-1)$

${\underline{\sf{Answer}}}$

$\underbrace{\bf{We\:Have}}$

• Initial Velocity = 36km/hr =10m/s
• Final velocity= 54km/hr = 15m/s
• Time ,t = 10sec

1) We have to find the acceleration of thd train

By using equation of motion:

$\sf\:v=u+at$

$\sf\:15=10+10a$

$\sf\:5=10a$

$\sf\:a=\dfrac{5}{10}$

$\sf\:a=\dfrac{1}{2}ms{}^{-2}$

2) We have to find the distance travelled by the train in 10 sec

By using equation of motion:

$\sf\:v{}^{2}=u{}^{2}+2as$

$\sf\:15{}^{2}=10{}^{2}+2\times\dfrac{1}{2}\times\:s$

$\sf\:225=100+s$

$\sf\:225-100=s$

$\sf\:s=125m$

Therefore,

• Acceleration= 1/2 m/s²
• And the distance travelled by the train is 125m