Question

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

Answer 1

initial velocity(u)=36km/hr=10m/s

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

Time=10 seconds.

Using first equation of motion to derive the value of Acceleration

v=u+at

15m=10+10a

5=10a

a=1/2m/s^2

Now,

we know the value of a

Using the equation 2 to find the distance

s=ut+1/2a(t)^2.

s=10×10+1/2×1/2×100

s=100+25m

S=125.

Thus,The distance travelled by train is 125m.

Answer 2

$\huge\tt{Answer:-}$

$\bf{Given:-}$

• Initial Velocity $(u)$ = $36 \ km \ {h}^{-1}$
• Final Velocity $(v)$ = $54 \ km \ {h}^{-1}$
• Time taken $(t)$ = $10 \ s$

$\bf{To \ Find:-}$

• Acceleration $(a)$.
• Distance traveled $(s)$.

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Case¹ : Acceleration:-

We know,

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

$\therefore a = \frac{(54 \ km \ {h}^{-1}) - (36 \ km \ {h}^{-1})}{10 \ s}$

$\implies a = \frac{54 \ km \ {h}^{-1} - 36 \ km \ {h}^{-1}}{10 \ s}$

$\implies a = \frac{18 \ km \ {h}^{-1}}{10 \ s}$

Convertion of (km m^-1) to (m s^-1):-

(Multiply the numerator with 5 and then divide it by 18)

$\implies a = \frac{5 \ m \ {s}^{-1}}{10 \ s}$

$\implies a = \frac{1}{2} \ m \ {s}^{-2}$

$\implies a = 0.5 \ m \ {s}^{-2}$ $...(Ans. \ 1)$

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Case² : Distance traveled:-

We know,

$\boxed{\tt{s = ut + \frac{1}{2}a {t}^{2}}}$

$\therefore s = (36 \ km \ {h}^{-1} \times 10 \ s) + [\frac{1}{2} \times 0.5 \ m \ {s}^{-2} \times {(10 \ s) }^{2}]$

$\implies s = (10 \ m \ {s}^{-1} \times 10 \ s) + (0.25 \ m \ {s}^{-2} \times 100 {s}^{2})$

(Convertion in similar manner)

$\implies s = (100 \ m) + (25 \ m)$

$\implies s = 125 \ m$ $...(Ans. \ 2)$