Answer:
total time taken is 2.1389 seconds
Explanation:
Given the data in the question and as illustrated in the diagram below;
They have measured the acceleration of the ball on the incline to be 2.5 m/s²
a = -2.5 m/s² { negative because its acting downward }
from the diagram, Motion from B to C will be;
v² = u² + 2as
given that distance s = 1.1 m
we substitute
v² - u² = 2as
(0)² - u² = 2 × (-2.5 m/s ) × 1.1 m
-u² = -5.5
u = √5.5
so Initial speed; u = 2.345 m/s
to get the time, we use the equation of motion;
v = u + at
we substitute
0 = 2.345 m/s + ( -2.5 m/s² × t₁ )
2.5t₁ = 2.345
t₁ = 2.345 / 2.5
t₁ = 0.938 s
Now, from C to B
we determine t₂
s = vt₂ + [tex]\frac{1}{2}[/tex]at₂²
1.1 m = 0×t₂ + [tex]\frac{1}{2}[/tex] × 2.5 m/s² × t₂²
1.1 m = [tex]\frac{1}{2}[/tex] × 2.5 m/s² × t₂²
1.1 m = 1.25 m/s² × t₂²
t₂² = 1.1 m / 1.25 m/s²
t₂² = 0.88 s²
t₂ = √0.88 s²
t₂ = 0.938 s
Next, from B to A; t₃ will be;
s = ut₃ + [tex]\frac{1}{2}[/tex]at₃²
we substitute
0.7 m = (2.345 m/s)t₃ + [tex]\frac{1}{2}[/tex] × 2.5 m/s × t₃²
0.7 m = (2.345 m/s)t₃ + ( 1.25 m/s )t₃²
1.25t₃² + 2.345t₃ - 0.7 = 0
using; x = [ -b ± √( b² - 4ac ) ] / 2a
we substitute
x = = [ -2.345 ± √( (2.345)² - (4 × 1.25 × -0.7) ) ] / 2(1.25)
x = [ -2.345 ± √( 5.499 + 3.5 ) ] / 2.5
x = [ -2.345 ± √( 8.999 ) ] / 2.5
x = [ -2.345 ± 2.9998 ] / 2.5
x = ( [ -2.345 - 2.9998 ] / 2.5 ) or ( [ -2.345 + 2.9998 ] / 2.5 )
x = -2.1379 or 0.2629
hence t₃ = 0.2629 s
∴ Total time = t₁ + t₂ + t₃
Total time = 0.938 s + 0.938 s + 0.2629 s
Total time = 2.1389 seconds
Therefore, total time taken is 2.1389 seconds
