Physics Motion in Straight Line

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New answer posted

2 months ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

Let height of bridge = h

Displacement of ball, S = - h

S = u t + 1 2 a t 2 - h = 4 * 4 + 1 2 ( - 10 ) ( 4 ) 2 h = 64 m

New answer posted

2 months ago

0 Follower 1 View

R
Raj Pandey

Contributor-Level 9

Average speed = 4 v 2 3 v

= 4 v 3

New answer posted

2 months ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

S nth   = u + a 2 ( 2 n - 1 ) S nth   ( 2 n - 1 ) = 1 : 3 : 5 : 7

New answer posted

2 months ago

0 Follower 3 Views

R
Raj Pandey

Contributor-Level 9

Slope of ( x - t )  graph, gives velocity

V 1 V 2 = t a n ? θ 1 t a n ? θ 2 = t a n ? 30 ? t a n ? 45 ? = 1 3

New answer posted

2 months ago

0 Follower 3 Views

R
Raj Pandey

Contributor-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

2 months ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-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

2 months ago

0 Follower 1 View

V
Vishal Baghel

Contributor-Level 10

Distance moved = Area under curve
= ½ * 8 * 5 = 20

New answer posted

2 months ago

0 Follower 5 Views

A
alok kumar singh

Contributor-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

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