Question

A train accelerates from 36 km/h to 54 km/h in 10 sec. Find:

Acceleration

The distance travelled by car.

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Answer 1

Q. A train accelerates from 36 km/h to 54 km/h in 10 sec. Find:-Acceleration and The distance travelled by car.

Ans :-

A = 0.5m/s²

S= 125 m

Given :-

U = 36 km/hr

change it into m/s

→36 × 5/18 = 10 m/s

V = 54 km/hr

change it in to m/s

→ 54 × 5/18 = 15 m/s

T = 10 sec.

Solution:-

Acceleration :- The rate of change of velocity is called acceleration.

Form definition,

$\boxed{\sf{ A = \dfrac {v-u}{t}}}$

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

$A = \dfrac {5}{10}$

$A = 0.5 m/s^2$

Now, distance travelled is given by

$2as = V^2 - u^2$

$2\times 0.5\times s = (15)^2-(10)^2$

$1 \times s = 225 - 100$

$s = 125 m$

hence, acceleration will be 0.5 m/s²

acceleration will be 0.5 m/s²and distance travelled will be 125 m.