The velocity of a car moving on a straight road increases linearly accroding to equation, `v=1+bx` , where `a &b` are positive constants. The acceleration in the course of such motion `:(x` is the displacement `)`

The speed v of a car moving on a straight road changes according to equation, v^2 = 1 + bx , where a and b are positive constants. Then the magnitude of acceleration in the course of such motion : (x is the distance travelled).

A particle moves in a straight line so that its velocity at any point is given by v^(2)=a+bx , where a,b ne 0 are constants. The acceleration is

A street car moves rectilinearly from station A (here car stops) to the next station B (here also car stops) with an acceleration varying according to the law f = a - bx , where a and b are positive constants and x is the distance from station A . If the maximum distance between the two stations is x = (Na)/(b) then find N .

If velocity of a particle is v=bs^(2) ,where b is constant and s is displacement. The acceleration of the particle as function of s is?

If the velocity v of particle moving along a straight line decreases linearly with its displacement s from 20 ms^(-1) to a value approaching zero at s = 30 m, then acceleration of the particle at v=10ms^(-1) is

If the velocity v of particle moving along a straight line decreases linearly with its displacement s from 20 ms^(-1) to a value approaching zero at s = 30 m, then acceleration of the particle at v=10ms^(-1) is

Figure below shows the velocity-time graph of a car moving on a straight road.The corresponding acceleration-=time graph will be: .

A street car moves rectilinearly from station A to the next station B with an acceleration varying according to the law f=a-bx , where a and b are constants and x is the distance from station A. The distance between the two stations & the maximum velocity are :