Physics Motion in Straight Line
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New answer posted
4 weeks agoContributor-Level 9
For constant acceleration, a,
x = ut + (1/2)at² ⇒ x-t graph is parabola.
v = u + at ⇒ v-t graph is straight line with positive slope.
New answer posted
4 weeks agoContributor-Level 10
t? + t? = t
v? /t? = α and v? /t? = β
t? = v? /α and t? = v? /β
v? /α + v? /β = t
v? (1/α + 1/β) = t
v? = (αβt) / (α + β)
Distance traveled = Area under speed-time graph
Distance = ½ * base * height = ½ * t * v?
Distance = ½ * t * (αβt) / (α + β) = (αβt²) / (2 (α + β)
New answer posted
a month agoContributor-Level 10
V? = 36 km/hr = 10 m/s
V? = -72 km/hr = -20 m/s
V? = -1.8 km/hr = -0.5 m/s
V? = V? + V?
= V? + V? - V?
= -0.5 + 10 – (-20)
= -0.5 + 30 = 29.5 m/s
New answer posted
a month agoContributor-Level 10
Area of OABS is A?
Area of SCD is A?
Distance = |A? | + |A? |
A? = (1/2) [13/3 + 1]4 = 32/3
A? = (1/2) * 5/3 * 2 = 5/3
Distance = |A? | + |A? |
= 32/3 + 5/3 = 37/3
New answer posted
a month agoContributor-Level 10
h = u²/2g, u = √2gh
Now, S = h/3
S = ut + ½at²
h/3 = √2ght - ½gt²
t² - 2√ (2h/g)t + 2h/3g = 0
Using quadratic formula for t:
t = ( 2√ (2h/g) ± √ (8h/g) - 4 (2h/3g) / 2
t = √ (2h/g) ± √ (2h/g - 2h/3g) = √ (2h/g) ± √ (4h/3g)
t? /t? = (√ (2h/g) - √ (4h/3g) / (√ (2h/g) + √ (4h/3g)
t? /t? = (√2 - √ (4/3) / (√2 + √ (4/3) = (√6 - 2)/ (√6 + 2)
(Note: There is a calculation error in the provided solution. Re-evaluating the physics.)
h/3 = (√2gh)t - ½gt²
(g/2)t² - (√2gh)t + h/3 = 0
t = (√2gh ± √ (2gh - 4 (g/2) (h/3)/g = (√2gh ± √ (4gh/3)/g
t? /t? = (√2gh - 2√gh/√3)/ (√2gh + 2√gh/√3)
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