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

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Given :-

U = 36 km/h = $36 \times\frac{1000}{60 \times 60}$ = 10 m/s

V = 54 km/h = $36 \times\frac{1000}{60 \times 60}$ = 15 m/s

T = 10 s

A = ?

S = ?

Let's find the 1 Acceleration :

A = $\frac{V - U}{T}$

A = $\frac{15 - 10}{10}$

A = 0.5 m/${s}^{2}$

Let's find the Distance :

S = UT + $\frac{1}{2}\:A {T}^{2}$

S = 10 + 10 $\frac{1}{2}×\:0.5× {10}^{2}$

S = 20 + 5

S = 25 m


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