Question

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

oh now how r u

Answer 1

Answer:

Acceleration =0.5m/a

Distance = 125m

Explanation:

Convert into SI units both the speed and apply equations of motion.

Answer 2

Given -

• $\:\:\:\tt{initial \:velocity, u = 36\: km/h}$
• $\:\:\:\tt{final \:velocity, v = 54\: km/h}$
• $\:\:\:\tt{time \:taken, t = 10 sec}$

To find -

(i) Acceleration

(ii) The distance travelled by car.

Solution -

Note -

• v = final velocity, $\:\:\:\:\:\tt{(m/s \:or\:ms^{-1})}$
• u = initial velocity, $\:\:\:\:\:\tt{(m/s\:or\:ms^{-1})}$
• t = time taken and $\:\:\:\:\:\tt{(s)}$
• a = acceleration $\:\:\:\:\:\tt{({m/s}^2\:or\:ms^{-2})}$

Firstly we'll convert the given velocities from km/h to m/s as the time is given in seconds,

$\:\:\:\dashrightarrow\sf{u=\{36\times\frac{5}{18}\}m/s \:or\:ms^{-1}}$

$\:\:\:\dashrightarrow\bf{u=10\:m/s}$

and

$\:\:\:\dashrightarrow\sf{v=\{54\times\frac{5}{18}\}m/s\:or\:ms^{-1}}$

$\:\:\:\dashrightarrow\bf{v=15\:m/s}$

(i) $\sf{Acceleration = \frac{v-u}{t}}$

$\:\:\:\implies\sf{Acceleration =\{ \frac{15-10}{10}\}m/s^2\:or\:ms^{-2}}$

$\:\:\:\implies\sf{Acceleration = \frac{5}{10}}$

$\:\:\:\implies{\boxed{\bf{Acceleration = 0.5\:m/s^2 \:or\: ms^{-2}}}}$

(ii) $\sf{Distance\: travelled=ut+\frac{1}{2}at^2}$

$\:\:\:\implies\sf{Distance\: travelled=\{(10)(10)+\frac{1}{2}\times\frac{5}{10}(10)^2\}m}$

$\:\:\:\implies\sf{Distance\: travelled=\{100+\frac{1}{\cancel{2}}\times\frac{5}{1\cancel{0}}\times\cancel{10}\cancel{0}\}m}$

$\:\:\:\implies{\boxed{\bf{Distance\: travelled=124m}}}$