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 <math><mrow><mfrac><mrow><msub><mrow><mi>t</mi></mrow><mrow><mn>1</mn></mrow></msub></mrow><mrow><msub><mrow><mi>t</mi></mrow><mrow><mn>2</mn></mrow></msub></mrow></mfrac></mrow></math> will be

tanα=vmaxt1=a1tanβ=vmaxt2=a2⇒a1a2=t2t1⇒t1t2=a2a1

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&nbsp;<math><mrow><mfrac><mrow><msub><mrow><mi>t</mi></mrow><mrow><mn>1</mn></mrow></msub></mrow><mrow><msub><mrow><mi>t</mi></mrow><mrow><mn>2</mn></mrow></msub></mrow></mfrac></mrow></math>&nbsp;will be

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

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

Option 1 -

$\frac{a_{1} + a_{2}}{a_{1}}$

Option 2 -

$\frac{a_{2}}{a_{1}}$

Option 3 -

$\frac{a_{1} + a_{2}}{a_{2}}$

Option 4 -

$\frac{a_{1}}{a_{2}}$

2 Views | Posted 2 months ago

Asked by Shiksha User

1 Answer

Answered by

Payal Gupta | Contributor-Level 10

2 months ago

Correct Option - 2

Detailed Solution:

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

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

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

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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] :

alok kumar singh

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

less

Payal Gupta

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 + $\frac{1}{2} a t^{2}$

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

80 = 2u + 5 × 2 × 2

80 = 2u + 20

2u = 60

u = 30m/sec

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)

alok kumar singh

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

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)

Vishal Baghel

h = 39.2 + 19.6 = 58.8 m

Height above tower

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

= 19.6

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

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

alok kumar singh

$K = \frac{(\frac{Q}{r}) Δ x}{Δ A T}$

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

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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