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Motion in a Straight Line

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3 months ago

Motion in a Straight Line Class 11th

Two spherical balls having equal masses with radius of 5cm each are thrown upwards along the same vertical direction at an interval of 3s with the same initial velocity of 35m/s, then these balls collide at a height of…………. m. (Take g = 10 m/s²)

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Vishal Baghel

Contributor-Level 10

y₁ = y₂

=> $35 t - \frac{1}{2} g t^{2} = 35 (t - 3) - \frac{1}{2} g {(t - 3)}^{2}$

=> 35 * 3 = $\frac{1}{2} g (t + t - 3) * 3$

105 = $\frac{10}{2} * 3 * (2 t - 3)$

t = 5

$y_{1} = 35 * 5 - \frac{1}{2} * 10 * 5^{2}$

= 50 m

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New answer posted

3 months ago

Motion in a Straight Line Class 11th

Two buses P AN Q start form a point at the same time and move in a straight line and their positions are represented by X_p (t) = at + bt² and X_Q (t) = ft – t². At what time, both the buses have same velocity?

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Raj Pandey

Contributor-Level 9

$X_{P} (t) = α t + β t^{2}$

$V_{P} (t) = α + 2 β t$ - (i)

$X_{Q} (t) = f t - t^{2}$

$V_{Q} (t) = f - 2 t$ - (ii)

(i) = (ii)

$α + 2 β t = f - 2 t \Rightarrow t (2 β + 2) = f - α$

$t = \frac{f - α}{2 (β + 1)}$

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New answer posted

3 months ago

Motion in a Straight Line Class 11th

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Raj Pandey

Contributor-Level 9

$X_{P} (t) = α t + β t^{2}$

$V_{P} (t) = α + 2 β t$ - (i)

$X_{Q} (t) = f t - t^{2}$

$V_{Q} (t) = f - 2 t$ - (ii)

(i) = (ii)

$α + 2 β t = f - 2 t \Rightarrow t (2 β + 2) = f - α$

$t = \frac{f - α}{2 (β + 1)}$

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New answer posted

3 months ago

Motion in a Straight Line Class 11th

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alok kumar singh

Contributor-Level 10

$X_{P} (t) = α t + β t^{2}$

$V_{P} (t) = α + 2 β t$ -(i)

$X_{Q} (t) = f t - t^{2}$

$V_{Q} (t) = f - 2 t$ -(ii)

(i) = (ii)

$α + 2 β t = f - 2 t \Rightarrow t (2 β + 2) = f - α$

$t = \frac{f - α}{2 (β + 1)}$ &

...more

$X_{P} (t) = α t + β t^{2}$

$V_{P} (t) = α + 2 β t$ -(i)

$X_{Q} (t) = f t - t^{2}$

$V_{Q} (t) = f - 2 t$ -(ii)

(i) = (ii)

$α + 2 β t = f - 2 t \Rightarrow t (2 β + 2) = f - α$

$t = \frac{f - α}{2 (β + 1)}$

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New answer posted

3 months ago

Motion in a Straight Line Class 11th

A NCC parade is going at a uniform speed of 9 km/h under a mango tree on which a monkey is sitting at a height of 19.6m. At any particular instant, the monkey drops a mango. A cadet will receive the mango whose distance from the tree at time of drop is:

(Given g = 9.8 m/s²)

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Vishal Baghel

Contributor-Level 10

$t = \sqrt[]{\frac{2 h}{g}} = \sqrt[]{\frac{2 * 19.6}{9.8}} = 2 s e c$

$∴ x = u t = \frac{9 * 1 0^{3}}{3600} * 2 = 5 m$

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New answer posted

3 months ago

Motion in a Straight Line Class 11th

From the top of a tower, a ball is thrown vertically upward which reaches the ground in 6s. A second ball thrown vertically downward from the same position with the same speed reaches the ground in 1.5 s. A third ball released, from the rest from the same location, will reach the ground in __________ s.

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Vishal Baghel

Contributor-Level 10

For first ball

s = ut + $\frac{1}{2} a t^{2}$

$\Rightarrow - h = u * 6 - \frac{1}{2} * 10 * 6 * 6$

$\Rightarrow - h = 6 u - 180$

h = 180 - 6u -(1)

for second boy

$h = u t \frac{1}{2} a t^{2}$

$h = u * 1.5 + \frac{1}{2} * 10 * 1.5 * 1.5$

h = 1.5u + 11.25 -(2)

from (1) & (2)

180 - 6u = 1.5u + 11.25

7.5u = 180 - 11.25

$u = \frac{168.75}{7.5} = 22.5$

h = 180 - 6u

= 180 - 6 * 22.5

45m

for third ball

$h = u t + \frac{1}{2} a t^{2}$

$h = 0 * t + \frac{1}{2} * 10 t^{2}$

$\Rightarrow 45 = 5 t^{2}$

t² = 9

t = 3 sec

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New answer posted

3 months ago

Motion in a Straight Line Physics NCERT Exemplar Solutions Class 12th Chapter Three

Two balls A and B placed at the top of 180m tall tower. Ball A is from the top at t = 0s. Ball B is thrown vertically down with an initial velocity 'u' at t = 2s. After a certain time, both balls meet 100m above the ground. Find the value of 'u' in ms^-¹.

[use g = 10 ms^-²] :

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

Let 't' be the time taken by B to meat A.

for A, u = 0

t = [2 + t]

h = 80m

a = 10 m/sec²

h = ut + $\frac{1}{2} a t^{2}$

80 = 0 * t + $\frac{1}{2} (10) {(2 + t)}^{2}$

16 = (2 + t)²

t + 2 = 4

t = 2sec

Now for B

h = ut + $\frac{1}{2} a t^{2}$

...more

Let 't' be the time taken by B to meat A.

for A, u = 0

t = [2 + t]

h = 80m

a = 10 m/sec²

h = ut + $\frac{1}{2} a t^{2}$

80 = 0 * t + $\frac{1}{2} (10) {(2 + t)}^{2}$

16 = (2 + t)²

t + 2 = 4

t = 2sec

Now for B

h = ut + $\frac{1}{2} a t^{2}$

$? 80 = u * 2 + \frac{1}{2} * 10 t^{2}$

80 = 2u + 5 * 2 * 2

80 = 2u + 20

2u = 60

u = 30m/sec

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New answer posted

3 months ago

Motion in a Straight Line Physics NCERT Exemplar Solutions Class 12th Chapter Three

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Payal Gupta

Contributor-Level 10

Let 't' be the time taken by B to meat

for A, u = 0

t = [2 + t]

h = 80m

a = 10 m/sec²

h = ut + $\frac{1}{2} a t^{2}$

80 = 0 * t + $\frac{1}{2} (10) {(2 + t)}^{2}$

16 = (2 + t)²

t + 2 = 4

t = 2sec

Now for B

h = ut + $\frac{1}{2} a t^{2}$

$\Rightarrow 80 = u * 2 + \frac{1}{2} * 10 t^{2}$

80 = 2u + 5 * 2 * 2

80 = 2u + 20

2u = 60

u = 30m/sec

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New answer posted

3 months ago

Motion in a Straight Line Physics NCERT Exemplar Solutions Class 12th Chapter Three

A scooter accelerates from rest for time t₁ at constant rate a₁ and then retards at constant rate a₂ for time t₂ and comes to rest. The correct value of $\frac{t_{1}}{t_{2}}$ will be

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

$t a n α = \frac{v_{m a x}}{t_{1}} = a_{1}$

$t a n β = \frac{v_{m a x}}{t_{2}} = a_{2}$

$\Rightarrow \frac{a_{1}}{a_{2}} = \frac{t_{2}}{t_{1}} \Rightarrow \frac{t_{1}}{t_{2}} = \frac{a_{2}}{a_{1}}$

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New answer posted

3 months ago

Motion in a Straight Line Physics NCERT Exemplar Solutions Class 12th Chapter Three

A Tennis ball is released from a height $h$ and after freely falling on a wooden floor it rebounds and reaches height $\frac{h}{2}$ . The velocity versus height of the ball during its motion may be represented graphically by:

(graphs are drawn schematically and on not to scale)

0 Follower 10 Views

alok kumar singh

Contributor-Level 10

$\begin{array}{r} \Rightarrow V^{2} = U^{2} + 2 g S \\ \Rightarrow V^{2} = 0 + 2 g (h - y) \\ \Rightarrow V^{2} = 2 g h - 2 g y \\ \Rightarrow V = \sqrt[]{2 g h - 2 g y} \end{array}$

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