Question

a body traveled a distance 2m in 2s and 2.2m in the next 2s. what will be the velocity at the end of the Start ​

Answer 1

The velocity at the end of the 7th second is 1.198 m/s

Explanation:

Assuming that the motion of the body is a uniformly accelerated motion (=constant acceleration), we can use the following suvat equation:

$s=ut+\frac{1}{2}at^2$

where

s is the displacement

u is the initial velocity

t is the time elapsed

a is the acceleration

In the first 2 seconds, the body travels 2 m, so we have:

$s_1 = 2 m\\t_1 = 2 s$

So the equation can be written as

$s_1=ut_1 + \frac{1}{2}at_1^2\\2=2u+2a$

If we then consider the whole first 4 seconds of motion, we have:

$s_2 = 2 + 2.2 = 4.2 m\\t_2 = 2+2 = 4 s$

So the equation can be written as

$s_2=ut_2+\frac{1}{2}at_2^2\\4.2 = 4u+8a$

So we have a system of 2 equations in 2 variables:

$2=2u+2a\\4.2=4u+8a$

By multiplying the 1st equation by 2 and subtracting eq.(1) from eq.(2), we get

$0.2=6a\\ \rightarrow a=\frac{0.2}{6}=0.033 m/s^2$

And susbtituting into the 1st equation, we get

$u=1-a=0.967 m/s$

Now we can apply the following suvat equation:

$v=u+at$

And substituting t = 7 s, we find the velocity at the end of the 7th second:

$v=0.967+(0.033)(7)=1.198 m/s$

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