Question

A train starting from rest attains a velocity of 108 km/h in 10 minutes
Find acceleration & distance

Answer 1

answer in the picture.....

Answer 2

The value of acceleration and  distance  is $0.05 \frac{m}{sec^{2} }$   and 9000 meter respectively.

Explanation:

Given:

A train starting from rest attains a velocity of 108 km/h in 10 minutes.

To Find:

The value of acceleration  and distance.

Formula Used:

The first equation of motion

$v=u+at$    --------------- formula no.01.

The third equation of motion

$v^{2} =u^{2} +2as$    ------------ formula no.02.

Where:

u =  the initial velocity

v = the final velocity

a= acceleration

t=time

s= displacement

Solution:

As given-a train starting from rest attains a velocity of 108 km/h in 10 minutes.

u=0   ,    $t =10\ minutes = 10\times60=600 \sec$

$v=108 \ km/h = \frac{108 \times1000}{60\times60} \ m/sec =30 \m/sec$

Applying formula no.01  and putting the value of u,v and t.

$v=u+at$

$30=0+a\times 600$

$30=a\times 600$

$a=0.05 \frac{m}{sec^{2} }$

Applying formula no.02 and putting the value of u,v and a.

$v^{2} =u^{2} +2as$

$30^{2} =0^{2} +2\times0.05\times s$

$900 =0+2\times0.05\times s$

$900 =0.1\times s$

$s=9000 \ meter$

Thus,the value of acceleration and  distance  is $0.05 \frac{m}{sec^{2} }$   and 9000 meter respectively.

PROJECT CODE #SPJ3