Question

$\bf{Physics!}$

Solve questions in the attachment!! ​

Answer 1

Second question solution :

Using the equation,

v = u + at

v = 0 + 5×4

v = 20 m/s

Distance travelled = vt = 20×4 = 80m

Answer 2

Question 1 :-

The average speed of the train is $\boxed{ \sf \red{19.4 m/s}}$

For the process kindly refer to the attachment(s) :)

Question 2 :-

★ Given :-

• u = 0
• a = 5 m/s²
• Time taken = 4 seconds

★ To Find :-

• Distance travelled after 4 seconds
• Speed after 4 seconds ( v )

★ Formula Used :-

In this question, we will use the two equations of motion, they are :-

$\boxed{ \sf \: 1. \: v = u + at}$

$\boxed{ \sf \: 2. \: s = ut + \frac{1}{2} a {t}^{2} }$

★ Solution :-

First of all we will find speed ( v) ,

$\sf : \implies \: v = 0 + 5 \times 4 \\ \sf : \implies \: v = 20 \: m/s \: \: \: \: \:$

Now, we will find out the distance covered in 4 seconds using the second equation of motion,

$\sf : \implies s = 0 \times 4 + \frac{1}{2} \times 5 \times ({4})^{2} \\ \sf : \implies s = \frac{1}{2} \times 5 \times 16 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf : \implies s = \frac{1} \orange{ \cancel2} \times 5 \times \green{ \cancel{16}} \: \:^{8} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf : \implies s =5 \times 8 = 40 \: m \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Hope that it helps you $\boxed{ \sf \: \orange{ @aanyaawesome008}}$ :)