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

tanα=vmaxt1=a1 tanβ=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 $\frac{t_{1}}{t_{2}}$ will be

Option 1 - <math> <mrow> <mfrac> <mrow> <msub> <mrow> <mi>a</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> <mo>+</mo> <msub> <mrow> <mi>a</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> </mrow> <mrow> <msub> <mrow> <mi>a</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> </mrow> </mfrac> </mrow> </math>

Option 2 - <math> <mrow> <mfrac> <mrow> <msub> <mrow> <mi>a</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> </mrow> <mrow> <msub> <mrow> <mi>a</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> </mrow> </mfrac> </mrow> </math>

Option 3 - <math> <mrow> <mfrac> <mrow> <msub> <mrow> <mi>a</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> <mo>+</mo> <msub> <mrow> <mi>a</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> </mrow> <mrow> <msub> <mrow> <mi>a</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> </mrow> </mfrac> </mrow> </math>

Option 4 - <math> <mrow> <mfrac> <mrow> <msub> <mrow> <mi>a</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> </mrow> <mrow> <msub> <mrow> <mi>a</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> </mrow> </mfrac> </mrow> </math>

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Payal Gupta | Contributor-Level 10

6 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. &n

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

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

80 = 2u + 5 × 2 × 2

80 = 2u + 20

2u = 60

u = 30m/sec

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(graphs are drawn schematically and on not to scale)

Alok Kumar Singh

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