The relation between time t and distance x is t=ax2 +bx where a and b are constants. The acceleration is

  • −2abv2

  • 2bv3

  • −2av3

  • 2av2


C.

−2av3

 if t = ax2+bx
then, differentiate both side with respect to time (t)
 
dtdt = 2axdxdt +bdxdt1 = 2axv +bvv.(2ax + b) = 1(2ax + b) = 1/v
 
Again differentiate both side with respect to time (t)
2adxdt = -v2.dvdt2av = - v-2.accelerationacceleration = -2av3
Hence, retardation= -2av3

A car travels along a straight line for first half time with speed 40 km/hr and the second half time with speed 60 km/hr. Find the mean speed of the car? 

Mean speed of car,


straight v subscript straight m space equals space fraction numerator straight v subscript 1 straight t subscript 1 space plus space straight v subscript 2 straight t subscript 2 over denominator straight t subscript 1 space plus space straight t subscript 2 end fraction space

space space space space space space equals space fraction numerator 40 space straight x space left parenthesis straight t divided by 2 right parenthesis space plus space 60 space left parenthesis straight t divided by 2 right parenthesis over denominator left parenthesis straight t divided by 2 space plus space straight t divided by 2 right parenthesis end fraction

space space space space space equals space 50 space km divided by hr

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The radius of the earth's orbit around the sun is 1.5 x 1011 m. Calculate the angular and linear velocity of the earth? 


Given, 

Radius of Earth's orbit, r = 1.5 x 1011

Time period of revolution of earth around the sun is 1 year. 

That is, 

T =  1 year = 365 x 24 x 60 x 60 s

Therefore, 

Angular velocity, straight omega space equals space fraction numerator 2 straight pi over denominator straight T end fraction

equals space fraction numerator 2 space cross times space open parentheses begin display style bevelled 22 over 7 end style close parentheses over denominator 365 space straight x space 24 space straight x space 60 space straight x 60 end fraction space equals space 1.99 space cross times space 10 to the power of negative 7 end exponent space sec

Linear velocity, v = straight omega space straight r 

                        equals space 1.99 space cross times space 10 to the power of negative 7 end exponent space cross times space 1.5 space cross times space 10 to the power of 11

equals space 2.99 space cross times space 10 to the power of 4 space straight m divided by straight s

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Two persons P and Q are standing 54 m apart on a long moving belt.  Person P rolls a round stone towards a person Q with a speed of 9 m/s with respect to belt. If the belt is moving with a speed of 4 m/s in the direction from P to Q. What is the speed of the stone w.r.to an observer on a stationary platform? 


Let the distance from P to Q be positive. 

Given, speed of the belt, vb = +4 m/s

Speed of the stone w.r.t belt, vs = + 9 m/s

Speed of the stone w.r.t a stationary observer= vs + vb = 9+ 4 = 13 m/s


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A plane flying horizontally at 100 m/s at a height of 1000 m releases a bomb from it. Find the time it takes to reach the ground. 

We have, 

Initial distance, yo = 0

Height attained after time t, y = 1000 m 

Initial velocity, uy = 0 m/s 

Acceleration, ay = 9.8 m/s2 

Time = t 

Now, using the equation of motion, we have 

space space space space straight y space equals space straight y subscript straight o space plus space straight u subscript straight y straight t space plus space 1 half straight a subscript straight y straight t squared space
therefore space 1000 space equals space 0 space plus space 0 space left parenthesis straight t right parenthesis space plus space 1 half cross times 9.8 cross times space straight t squared space

space space space space space space space space space space space space equals space 4.9 space straight t squared space

rightwards double arrow space straight t space equals space square root of fraction numerator 1000 over denominator 4.9 end fraction end root space equals space 100 over 7 space equals space 14.28 space sec

Therefore, 14.28 sec ia taken for the bomb to reach the ground. 

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