The equation of movement for this case is given by:
[tex] vf = at + vo [/tex]
Where,
vf: final speed
a: acceleration
t: time
vo: initial speed
Substituting values we have:
[tex] 6.18 = a (3.56) +6.98 [/tex]
Clearing the acceleration we have:
[tex] a = \frac{6.18-6.98}{3.56} [/tex]
[tex] a = -0.225 \frac{m}{s ^ 2} [/tex]
Answer:
his acceleration during this time is:
[tex] a = -0.225 \frac{m}{s ^ 2} [/tex]
