Answer:
Acceleration of the object is 4.11 [tex]\frac{m}{s^{2} }[/tex].
Given:
Initial velocity = 10 [tex]\frac{m}{s}[/tex]
Time = 6 second
Distance = 134 meter
To find:
Acceleration of the object = ?
Formula used:
s = u t + [tex]\frac{1}{2} a t^{2}[/tex]
Where s = distance covered by object
u = initial velocity of the object
t = time taken by object
a = acceleration of the object
Solution:
According to second equation of motion the distance of the object is given by,
s = u t + [tex]\frac{1}{2} a t^{2}[/tex]
Where s = distance covered by object
u = initial velocity of the object
t = time taken by object
a = acceleration of the object
134 = 10 (6) + [tex]\frac{1}{2} a (6)^{2}[/tex]
134 - 60 = 18 a
74 = 18 a
a = [tex]\frac{74}{18}[/tex]
a = 4.11 [tex]\frac{m}{s^{2} }[/tex]
Hence, acceleration of the object is 4.11 [tex]\frac{m}{s^{2} }[/tex].
