Some students are making a domino cascade for a fair project and want to use a ball rollingup and then down an incline in one part of it. The timing has to be very precise so theyneed to know exactly how long it will take the ball to go up and down. They have measuredthe acceleration of the ball on the incline to be 2.5 m/s2. If the ball starts part way up theramp, rolls up for 1.1 m, then down 0.7 m past its starting point, how long will it take

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 + at₂²

1.1 m = 0×t₂ + × 2.5 m/s² × t₂²

1.1 m = × 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 + at₃²

we substitute

0.7 m = (2.345 m/s)t₃ + × 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