Physics Motion in Straight Line

Get insights from 59 questions on Physics Motion in Straight Line, answered by students, alumni, and experts. You may also ask and answer any question you like about Physics Motion in Straight Line

Follow Ask Question
59

Questions

0

Discussions

6

Active Users

0

Followers

New answer posted

4 weeks ago

0 Follower 2 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

4 weeks 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

4 weeks ago

0 Follower 2 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

a month ago

0 Follower 1 View

V
Vishal Baghel

Contributor-Level 10

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

New answer posted

a month ago

0 Follower 4 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

New answer posted

a month ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-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 ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

Kindly consider the following Image

 

New answer posted

a month ago

0 Follower 6 Views

V
Vishal Baghel

Contributor-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)

...more

Get authentic answers from experts, students and alumni that you won't find anywhere else

Sign Up on Shiksha

On Shiksha, get access to

  • 65k Colleges
  • 1.2k Exams
  • 688k Reviews
  • 1800k Answers

Share Your College Life Experience

×

This website uses Cookies and related technologies for the site to function correctly and securely, improve & personalise your browsing experience, analyse traffic, and support our marketing efforts and serve the Core Purpose. By continuing to browse the site, you agree to Privacy Policy and Cookie Policy.