Answer:
2.58 m/s
Explanation:
We can use the following equation of motion to find out the constant acceleration of the block:
[tex]v^2 - v_0^2 = 2a\Delta s[/tex]
where v = 3.8 m/s is the final velocity after it has traveled 7.4 m, [tex]v_0 = 0[/tex] is the initial velocity of the block when it starts from rest, , and [tex]\Delta s = 7.4[/tex] is the distance traveled.
[tex]3.8^2 - 0 = 2*a*7.4[/tex]
[tex]14.44 = 14.8a[/tex]
[tex]a = 14.44 / 14.8 = 0.98 m/s^2[/tex]
We can use the same motion equation to calculate block speed at the end of 3.4m track
[tex]v_2^2 - 0 = 2*0.98*3.4[/tex]
[tex]v_2^2 = 6.63[/tex]
[tex]v_2 = \sqrt{6.63} = 2.58 m/s[/tex]
The speed of the block after 3.4 m is 2.58 m/s. Speed can be defined as the change in the position of an object over time.
Speed can be defined as the change in the position of an object over time.
First, calculate the acceleration,
[tex]V_f ^2- v_i^2 = 2 a \Delta s\\[/tex]
Where,
[tex]v_f[/tex] - final speed = 3.80 m/s
[tex]v_i[/tex]- initial speed = 0
[tex]\Delta s[/tex] - Distance traveled = 7.4 mm
[tex]a[/tex] - acclelearation = ?
Put the values in the formula,
[tex]3.8 - 0 = 2 \times a \times 7.4\\\\a = 0.98\rm \ m/s^2[/tex]
Put the value of acceleration,
[tex]v_f^2 - 0 = 2 \times a \times 3.4 \\\\v_f =2.58\rm \ m/s[/tex]
Therefore, the speed of the block after 3.4 m is 2.58 m/s.
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