Question

A train accelerate from 36 km/h to 54 km/h in 10 sec. Find acceleration and distance travelled by car
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Answer 1

Answer:

$\boxed{\mathfrak{Acceleration \ (a) = 0.5 \ m/s^2}}$

$\boxed{\mathfrak{Distance \ travelled \ (s) = 125 \ m}}$

Given:

Initial velocity (u) = 36 km/h = 10 m/s

Final velocity (v) = 54 km/h = 15 m/s

Time taken (t) = 10 seconds

To Find:

(i) Acceleration (a)

(ii) Distance travelled (s)

Explanation:

$\star$ Converting km/h to m/s:

$\sf 1 \:km/h = \frac{5}{18} \: m/s$

So,

$\sf Initial \: velocity \: (u) = 36 \: km/h \\ \\ \sf = 36 \times \frac{5}{18} \: m/s \\ \\ \sf = 2 \times 5 \: m/s \\ \\ \sf = 10 \: m/s \\ \\ \\ \sf Final \: velocity \: (v) = 54 \: km/h \\ \\ \sf = 54 \times \frac{5}{18} \: m/s \\ \\ \sf = 3 \times 5 \: m/s \\ \\ \sf = 15 \: m/s$

$\sf (i) \: From \: 1^{st} \: equation \: of \: motion: \\ \boxed{ \bold{v = u + at}}$

Substituting value of v, u & t in the equation:

$\sf \implies 15 = 10 + a(10) \\ \\ \sf \implies 10a + 10 = 15 \\ \\ \sf \implies 10a = 15 - 10 \\ \\ \sf \implies 10a = 5 \\ \\ \sf \implies a = \frac{5}{10} \\ \\ \sf \implies a = \frac{1}{2} \\ \\ \sf \implies a = 0.5 \: m/s$

$\therefore$

Acceleration (a) = 0.5 m/s

$\sf (ii) \: From \: 2^{nd} \: equation \: of \: motion: \\ \boxed{ \bold{s = ut + a {t}^{2} }}$

Substituting value of u, t & a in the equation:

$\sf \implies s = 10(10) + \frac{1}{2} \times 0.5 \times {(10)}^{2} \\ \\ \sf \implies s = 100 + \frac{1}{2} \times 0.5 \times 100 \\ \\ \sf \implies s = 100 + 0.5 \times 50 \\ \\ \sf \implies s = 100 + 25 \\ \\ \sf \implies s = 125 \: m$

$\therefore$

Distance travelled (s) = 125 m

Answer 2

Given :

• Initial velocity , u = 36km/hr =10m/s
• Final velocity ,v = 54 kmhr= 15m/s
• t= 10sec

To Find :

• Acceleration
• Distance travelled

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

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

Solution :

Part -1

We have to find the acceleration.

By using equation of motion

$\sf\:v=u+at$

Now put the given values

$\sf\:15=10+a\times10$

$\sf\:5=10a$

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

$\sf\:a=0.5ms^{-2}$

Hence ,Acceleration is 0.5m/s²

Part -2

We have to find the distance covered .

By using equation of motion

$\sf\:v^2=u^2+2aS$

Now put the given values

$\sf\:(15)^2=(10)^2+2\times0.5\times\:S$

$\sf\:225=100+2\times0.5\times\:S$

$\sf\:S=\dfrac{225-100}{0.5\times2}$

$\sf\:S=\dfrac{125}{2\times0.5}$

$\sf\:S=\dfrac{125\times10}{2\times5}$

$\sf\:S=125m$

Hence ,the distance travelled is 125m.