Home
Class 11
PHYSICS
A point P moves in counter - clockwise ...

A point `P` moves in counter - clockwise direction on a circular path as shown in the figure . The movement of `P` is such that it sweeps out a length `s = t^(3) + 5 ` , where `s` is in metres and ` t` is in seconds . The radius of the path is `20 m` . The acceleration of `P` when ` t = 2 s` is nearly .

A

`14 m//s^(2)`

B

`13 m//s^(2)`

C

`12 m//s^(2)`

D

`7.2 m//s^(2)`

Text Solution

Verified by Experts

The correct Answer is:
A
Here, `s = t^(3) + 5, v = (ds)/(dt) = 3t^(2), r = 20m`
Tangential acceleration,
`a_(T) = (d v)/(dt) = (d)/(dt)(3t^(2)) = 6t`
at `t = 2 sec, a_(T) = 6 xx 2 = 12 m//s^(2)`
Centripetal acceleration
`a_(c) = (v^(2))/(r ) = ((12)^(2))/(20) = (144)/(20)m//s^(2)`
Total acceleration,
`a = sqrt(a_(T)^(2) + a_(C)^(2)) = sqrt((2)^(2) + ((144)/(20))^(2)) ~~ 14 m//s^(2)`
Promotional Banner

Similar Questions

Explore conceptually related problems

A point P moves in counter clockwise direction on a circular path as shown in the figure. The movement of 'P' is such that it sweeps out a length s=t^(2)+5 , wheres is in metres and t is in seconds. The radius of the path is 20m. The acceleration of 'P' when t=5sqrt((3)/(10)) seconds is nearly:

A point P moves in a counter-clockwise direction on a circular path as shown in the figure. The movement of P is such that it sweeps out a length s= t^(3) +5 where s is in the metre and t is in seconds. The radius of the path is 27 m. The acceleration of P when t= 3 s is _________ m//s^(2) . (Take sqrt(13) = 3.6 )

A point P moves in counter-clockwise direction on figure. The movement of P is such that it sweeps out a length s=t^(3)+5 , where s is in metre and t is in second. The radius of the pathh is 20 m. the acceleration of P when t=2s is nearly

A point p moves in counter - clockwise direction on a circular path as shown in the figure . The movement of 'p' is such that it sweeps out in the figure . The movement of 'p' is such that it sweeps out a length s = t^(3) + 5 , where s is in metres and t is in seconds . The radius of the path is 20 m . The acceleration of 'P' when t = 2 s is nearly .

The displacement of a particle moving in a straight line, is given by s = 2t^2 + 2t + 4 where s is in metres and t in seconds. The acceleration of the particle is.

The displacement of a particle is represented by the equation, s=3 t^3 =7 t^2 +4 t+8 where (s) is in metres and (t) in seconds . The acceleration of the particle at t=1 s is.

A car is moving in such a ways that the distance it covers , is given by the equations s = 4t^(2) +3t, where s, is in meters and t is in seconds . What would be the velocity and the accelerations of the car at time t=20 seconds?

The displacment of a particle is represented by the following equation : s=3t^(3)+7t^(2)+5t+8 where s is in metre and t in second. The acceleration of the particle at t=1:-

A car moves along a straight line whose equation of motion is given by s=12t+3t^(2)-2t^(3) where s is in metres and t is in seconds. The velocity of the car at start will be :-