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

Explanation:

a = v - u /t

a = ?

v= 54km

u= 36km

t = 10s

a = 54 - 36 / 10

= 18/ 10

= 1.8

acceleration = 1.8

(ii) distance = 54km -36km

(s) = 18 km

Answer 2

Answer:

Velocity is the displacement covered by a body per unit time.

Acceleration is the rate of change of velocity with respect to time.

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Given initial velocity (u) = 36kmh¯1

Given final velocity (v) = 54kmh¯¹

Given time = 10 seconds

$(i) \: Acceleration = \frac{change \: in \: velocity}{total \: time}$

$= > \frac{final \: velocity - initial \: velocity}{time}$

$= > \frac{v - u}{t}$

$= > \frac{54 - 36}{10}$

$= > \frac{18}{10} = 1.8 \: m {s}^{ - 2}$

$(ii) Let \: Distance \: travelled \: by \: car \: be \: s.$

By using third equation of motion i.e 2as =v²-u².

$2 \times 1.8 \times s = {54}^{2} - {36}^{2}$

$3.6 \times s = 1620 \\ s = \frac{1620}{3.6} = 450 \: km$