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Motion in a Straight Line

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3 months ago

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

A ball is thrown vertically upwards with a velocity of 19.6 ms^-1 from the top of a tower. The ball strikes the ground after 6s. The height from the ground up to which the ball can rise will be $(\frac{k}{5}) m .$ The value of k is_______(use g = 9.8m/s²)

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Vishal Baghel

Contributor-Level 10

h = 39.2 + 19.6 = 58.8 m

Height above tower

$h' = \frac{u^{2}}{2 g} = \frac{19.6 * 19.6}{2 * 9.8}$

= 19.6

$A s, \frac{k}{5} = 78.4 \Rightarrow k = 392$

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3 months ago

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

Dimensional formula for thermal conductivity is (here $K$ denotes the temperature):

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alok kumar singh

Contributor-Level 10

$K = \frac{(\frac{Q}{r}) Δ x}{Δ A T}$

$\begin{array}{l} \Rightarrow & \frac{({M L}^{2} T^{- 2}) (L)}{(L^{2}) (θ) (T)} \\ \Rightarrow & M^{1} L^{1 -} T^{- 3} θ^{- 1} \end{array}$

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3 months ago

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

The speed verses time graph for a particle is shown in the figure. The distance travelled (in $m$ ) by the particle during the time interval $t = 0$ to $t = 5 s$ will be $\dots .$ .

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alok kumar singh

Contributor-Level 10

Sol.

$\Rightarrow n_{1} T_{1} + n_{2} T_{2} = n T$

$\Rightarrow (0.1) (200) + (0.05) (400) = (0.15) T$

$\Rightarrow T = 266.67$

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3 months ago

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

If t = $\sqrt[]{x} + 4, t h e n {(\frac{d x}{d t})}_{t = 4} i s :$

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Vishal Baghel

Contributor-Level 10

$t = \sqrt[]{x} + 4$

then $\frac{d t}{d x} = \frac{1}{2 \sqrt[]{x}} ∴ {(\frac{d x}{d t})}_{t = 4} = {(2 \sqrt[]{x})}_{t = 4} = {[2 (t - 4)]}_{a t t = 4}$

= 0

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New answer posted

3 months ago

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

A ball is thrown up vertically with a certain velocity so that, it reaches a maximum height h. Find the ratio of the times in which it is at height $\frac{h}{3}$ while going up and coming down respectively.

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Vishal Baghel

Contributor-Level 10

$h = \frac{u^{2}}{2 g} \Rightarrow u = \sqrt[]{2 g h}$

$\frac{h}{3} = \sqrt[]{2 g h} t - 5 t^{2}$

$5 t^{2} - \sqrt[]{20 h} t + \frac{h}{3} = 0$

$t = \frac{\sqrt[]{20 h} \pm \sqrt[]{20 h - 4 * 5 * \frac{h}{3}}}{10}$

$∴ \frac{t_{1}}{t_{2}} = \frac{\sqrt[]{20 h} - \sqrt[]{\frac{40}{3} h}}{\sqrt[]{20 h} + \sqrt[]{\frac{40}{3} h}} = \frac{1 - \sqrt[]{\frac{2}{3}}}{1 + \sqrt[]{\frac{2}{3}}} = \frac{\sqrt[]{3} - \sqrt[]{2}}{\sqrt[]{3} + \sqrt[]{2}}$

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New answer posted

3 months ago

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

If t = $\sqrt[]{x} + 4, t h e n {(\frac{d x}{d t})}_{t = 4} i s :$

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Vishal Baghel

Contributor-Level 10

$t = \sqrt[]{x} + 4$

then $\frac{d t}{d x} = \frac{1}{2 \sqrt[]{x}} ∴ {(\frac{d x}{d t})}_{t = 4} = {(2 \sqrt[]{x})}_{t = 4} = {[2 (t - 4)]}_{a t t = 4}$

= 0

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New answer posted

3 months ago

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

When a ball is dropped into a lake form a height 4.9 m above the water level, it hits the water with a velocity v and then sinks to the bottom with the constant velocity v. It reaches the bottom of the lake 4.0 s after it is dropped. The approximate depth of the lake is

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Payal Gupta

Contributor-Level 10

Total time (T) = 4 sec

Given u = 0 m/sec

a = g = 9.8 m/sec²

h = 4.9 m, t =?

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

$4.9 = 0 . t + \frac{1}{2} * 9.8 t^{2}$

t = 1 sec

'v' be the velocity with which ball hits the water v = u + at

= 0 + 9.8 * 1 = 9.8 m/sec

Time taken to reach the bottom of the lake from surface of the lake

= 4 – 1 = 3 sec

v = 9.8 m/sec

$H = u t + \frac{1}{2} g t^{2} = 9.8 * 3 + \frac{1}{2} * 9.8 * 9$

29.4 + 4.9 * 9 = 29.4 + 44.1

H = 73.5 m

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4 months ago

Motion in a Straight Line Physics NCERT Exemplar Solutions Class 11th Chapter Three

A ball is bouncing elastically with a speed 1 m/s between walls of a railway compartment of size 10 m in a direction perpendicular to walls. The train is moving at a constant velocity of 10 m/s parallel to the direction of motion of the ball. As seen from the ground,

(a) The direction of motion of the ball changes every 10 seconds.

(b) Speed of ball changes every 10 seconds.

(c) Average speed of ball over any 20 second interval is fixed.

(d) The acceleration of ball is the same as from the train.

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(b, c, d) velocity of ball before collision =10+1=11m/s

Speed after collision= 10-1=9m/s

As the speed is changing after travelling 10 m and speed is 1m/s hence, time duration of the changing speed is 10

Since the collision of the ball is perfectly elastic there is no dissipation of energy hence, total momentum and kinetic energy are conserved.

Since the train is moving with constant velocity hence, it will act as inertial frame of reference as that of earth and acceleration will be same in both cases.

As the collision is perfectly elastic so momentum and total energy (K.E and

...more

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New answer posted

4 months ago

Motion in a Straight Line Physics NCERT Exemplar Solutions Class 11th Chapter Three

A spring with one end attached to a mass and the other to a rigid support is stretched and released.

(a) Magnitude of acceleration, when just released is maximum.

(b) Magnitude of acceleration, when at equilibrium position, is maximum.

(c) Speed is maximum when mass is at equilibrium position.

(d) Magnitude of displacement is always maximum whenever speed is minimum.

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(a, c) as we know restoring force is F=-kx

Potential energy of the spring = 1/2kx²

The restoring force is central hence, when particle released it will execute SHM about equilibrium position

Also acceleration= f/m=-kx/m

But at equilibrium position when x=o, a also 0 and at equilibrium position potential energy converted into kinetic energy so speed will also maximum that time.

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New answer posted

4 months ago

Motion in a Straight Line Physics NCERT Exemplar Solutions Class 11th Chapter Three

For the one-dimensional motion, described by x = t–sint

(a) x (t) > 0 for all t > 0.

(b) v (t) > 0 for all t > 0.

(c) a (t) > 0 for all t > 0.

(d) v (t) lies between 0 and 2.

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(a, d) x= t-sint

Velocity v = dx/dt=

d/dt (t-sint)=

=1-cost

When cost =1, velocity v=0

V_max=1-cost_min = 1- (-1)=2

V_{min =}1- (cost)_max= 1-1=0

Hence v lies between 0 and 2

Acceleration a=dv/dt=-sint

When v=0 then cost =1

V_max= 1- (-1)=2, V_min=1-1=0

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