Question

A train starts from rest with an acceleration of 2m/s² . A man who is 9m behind the train runs with uniform speed and just manages to get into the train in 10 sec . The speed of the man is


pls answer quickly with step by step explaination​

Answer 1

Explanation:

u=0

a=2 m/s^2

t=10 sec

distance covered by train

s=ut+1/2at^2

=0*10+1/2*2*10^2

=0+100

=100m

now the man is 9 m behind

total distance =100+9=109

speed of man =distance /time

=109/10=10.9 m/s

I think it is helpful for you

Answer 2

$\dag \: \underline{\sf AnsWer :}$

In the question it is said that, a train starts from rest therefore initial velocity (u) will be 0 m/s and the trains moves with an acceleration (a) of 2 m/s². We are also given that a man who is 9m behind the train runs with uniform speed [Distance travelled by the man = 9 m] and time taken (t) by the to reach the train is 10 second. and we are asked to find the speed of man.

• First we need to find the distance travelled by the train by using second kinematical equation of motion :

$:\implies \sf s = ut + \dfrac{1}{2} at^2$

$:\implies \sf s = 0 \times 10 + \dfrac{1}{2} \times 2 \times (10)^2$

$:\implies \sf s = \dfrac{1}{2} \times 2 \times 100$

$:\implies \underline{ \boxed{\sf s = 100 \: m}}$

• Now we have find the distance travelled by the train that is 100 m and we are given the distance travelled by the man that is 9 m. So, now we can find the total distance travelled by the both train and man.

$\dashrightarrow\:\:\textsf {Total distance = Distance travelled by the train + Distance travelled by the man }$

$\dashrightarrow\:\:\textsf {Total distance = 100 + 9 }$

$\dashrightarrow\:\: \underline{ \boxed{\textsf {Total distance = 109 m}}}$

• Here we have find the total distance. Therefore, we can find the speed of man. So, let's do it :

$\longrightarrow\:\:\sf Speed \: of \: man = \dfrac{Total \: Distance}{Time}$

$\longrightarrow\:\:\sf Speed \: of \: man = \dfrac{109}{10}$

$\longrightarrow\:\: \underline{ \boxed{\sf Speed \: of \: man = 10.9 \: m/s}}$