Question

A train starting from rest attains a velocity of 72 km h -¹ in 5 minutes assuming that the acceleration is uniform find the acceleration and the distance travelled by the train for attaining this velocity

Answer 1

Hey friend
here is your answer

initial velocity = 0 m/sec
final velocity = 72 km/hr or 20 m/sec
time = 300 sec
so,
acceleration must be
$v = u + at \\ 20 = 0 + 300 \times a \\ \frac{1}{15}m {sec}^{ - 2} = a$

now
$s = ut + \frac{1}{2} a {t}^{2} \\ s = 0 \times 300 + \frac{1}{15} \times 300 \times 300 \\ s = 300 + 6000 \\ s = 6300m$
so, total distance travelled by him is 6300 m


glad to help you
hope it helps
thank you.

Answer 2

Answer:

$\frac{2}{30} \ m/s^{2}$, 3000 metres

Explanation:

Given:

• Initial velocity = u = 0 m/s
• Final velocity = v = 72 km/h = $72 \times\frac{5}{18}$=20 m/s
• Time = t = 5 minutes = 300 seconds

To find:

• Acceleration of the train
• Distance travelled by the train

Using first equation of motion:

V=u+at

20=0+a×300

20=300a

a = $\frac{2}{30} \ m/s^{2}$

Now using third equation of motion:

V²-u²=2as

20²-0²=2×$\frac{2}{30}$×s

400=$\frac{4}{30} s$

s = $400 \times \frac{30}{4}$

s = 3000 metres

The acceleration is $\frac{2}{30} \ m/s^{2}$ and distance covered is 3000 metres