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 3 months ago

Asked by Shiksha User

1 Answer

Answered by

Payal Gupta | Contributor-Level 10

3 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 t1t2 will be

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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