Question

a train starting from rest attains a velocity of 72 kilometre per hour in 5 minutes assuming that the acceleration is uniform find the acceleration and the distance travelled by the train for attending the velocity​

Answer 1

U=0

V=72km/h

T=5m

A=?

S(distance)=?

1. Equation of Motion : v=u+at. (To find accleration)

5

72km=72 * — = 30m/s

18

5m=5*60=300

30=0+A*300

30=300A

300A=30

30

A = —–

300

A=10m/s²

2. Equation of motion : v²=u²+2as. (To find distance)

30²=0+2*10*s

900=0+20s

20s=900

900

s = ——

20

s=45m

Answer 2

Question :

A train starting from rest attains a velocity of 72 kilometre per hour in 5 minutes assuming that the acceleration is uniform, find the acceleration and the distance travelled by the train for attending the velocity ?

Answer :

Given,

initial velocity (u) = 0 ;

final velocity (v) = 72 km/hr ;

$72 \times \frac{5}{18} \\ 4 \times 5$

20 m/s

Time (t) = 5min = 300 s

Now,

1. Acceleration =

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

$a = \frac{20 - 0}{300} \\ \frac{1}{15}m \: s {}^{ - 1}$

2. Distance =

From equation

$2 \: a \: s = v {}^{2} - u {}^{2} \\ v {}^{2} - 0$

Thus,

$s = \frac{v {}^{2} }{2a} \\ \frac{20}{2 \times 1 \div 15}$

3000 m

Change m in km

3 km

The acceleration of the train is 1/15 metre per second and the distance travelled in 3 km