Physics Motion in Straight Line

The velocity (v) and time (t) graph of a body in a straight line motion is shown in the figure. The point S is at 4.333 seconds. The total distance covered by the body in 6 s is:

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Contributor-Level 10

Area of OABS is A?

Area of SCD is A?
Distance = |A? | + |A? |
A? = (1/2) [13/3 + 1]4 = 32/3
A? = (1/2) * 5/3 * 2 = 5/3
Distance = |A? | + |A? |
= 32/3 + 5/3 = 37/3

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2 months ago

Physics Motion in Straight Line Motion in a Straight Line

The velocity- displacement graph describing the motion of abicycle is shown in the figure. The acceleration displacement graph of the bicycle's motion is best described by:

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Raj Pandey

Contributor-Level 9

Kindly consider the following Image

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2 months ago

A ball is thrown up with a certain velocity so that it reaches a height 'h'. Find the ratio of the two different times of the ball reaching h/3 in both the directions.

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Contributor-Level 10

h = u²/2g, u = √2gh
Now, S = h/3
S = ut + ½at²
h/3 = √2ght - ½gt²
t² - 2√ (2h/g)t + 2h/3g = 0
Using quadratic formula for t:
t = ( 2√ (2h/g) ± √ (8h/g) - 4 (2h/3g) / 2
t = √ (2h/g) ± √ (2h/g - 2h/3g) = √ (2h/g) ± √ (4h/3g)
t? /t? = (√ (2h/g) - √ (4h/3g) / (√ (2h/g) + √ (4h/3g)
t? /t? = (√2 - √ (4/3) / (√2 + √ (4/3) = (√6 - 2)/ (√6 + 2)
(Note: There is a calculation error in the provided solution. Re-evaluating the physics.)
h/3 = (√2gh)t - ½gt²
(g/2)t² - (√2gh)t + h/3 = 0
t = (√2gh ± √ (2gh - 4 (g/2) (h/3)/g = (√2gh ± √ (4gh/3)/g
t? /t? = (√2gh - 2√gh/√3)/ (√2gh + 2√gh/√3)

...more

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2 months ago

The instantaneous velocity of a particle moving in a straight line is given as v = αt + βt², where α and β are constants. The distance travelled by the particle between 1s and 2s is :

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Contributor-Level 10

v =? t +? t²
ds/dt =? t +? t²
? this =? (? t +? t²)dt from 1 to 2
s = [? t²/2 +? t³/3] from 1 to 2
s = [? (2)²/2 +? (2)³/3] - [? (1)²/2 +? (1)³/3]
s = [2? + 8? /3] - [? /2 +? /3]
s = 3? /2 + 7? /3

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2 months ago

The relation between time t and distance x for a moving body is given as t = mx² + nx, where m and n are constants. The retardation of the motion is: (Where v stands for velocity)

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Contributor-Level 10

By t = mx² + nx
dt/dx = 2mx + n
V = dx/dt = 1/ (2mx + n)
a = v (dv/dx) = V (d/dx (1/ (2mx+n) = V [-2m/ (2mx+n)²] = -2mV³
So, Retardation will be (2mv³)

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2 months ago

Water droplets are coming from an open tap at a particular rate. The spacing between a droplet observed at 4? second after its fall to the next droplet is 34.3 m. At what rate the droplets are coming from the tap? (Take g = 9.8 m/s²)

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Raj Pandey

Contributor-Level 9

Displacement in 4? second
= S? - S?
= (1/2)g [2n - 1]
= (1/2)g [2*4 - 1]
= (1/2) * 9.8 * 7
= 34.3m
As this distance matches with data given in question for position of next drop. So drops are falling at the rate of 1 drop/second.

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2 months ago

A balloon was moving upwards with a uniform velocity of 10 m/s. An object of finite mass is dropped from the balloon when it was at a height of 75 m from the ground level. The height of the balloon from the ground when object strikes the ground was around :
(takes the value of g as 10 m/s²)

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Contributor-Level 10

Body is dropped from height 75m with initial velocity (upward) 10m/s
So, s = ut + ½ at²
-75 = 10t - ½ gt²
5t² - 10t - 75 = 0
By solving t = 5 sec.
In 5sec. balloon covered
h = 10 * 5 = 50m
Now height of balloon = 75 + 50 = 125m

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2 months ago

If the velocity-time graph has the shape AMB, what would be the shape of the corresponding acceleration-time graph?

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Contributor-Level 10

For part AM, slope of v – t graph is constant but negative. For part MB, slope of v – t graph is constant but positive.

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3 months ago

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:

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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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3 months ago

An engine of a train, moving with uniform acceleration, passes the signal – post with velocity u and the last compartment with velocity $υ .$ The velocity with which middle point of the train passes the signal post is:

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