The acceleration of a particle is increasing linearly with time t as bt. The particle starts from origin with initial velocity u. The distance travelled by the particle in time t will be:

a) ut+(1/3)bt2

b) ut+(1/2)bt2

c) ut+(1/6)bt3

d) ut+(1/3)bt3

Question

The acceleration of a particle is increasing linearly with time t as bt. The particle starts from origin with initial velocity u. The distance travelled by the particle in time t will be:

a) ut+(1/3)bt2

b) ut+(1/2)bt2

c) ut+(1/6)bt3

d) ut+(1/3)bt3

Vikas TU · Answer

Displacement is the integral of accelration twice if the relation is gien in time. Therefore, a = bt (given) dv/dt = bt dv = bt dt integerating both sides we get, v = bt^2/2 dx = bt^2/2 dt x = bt^3/6 Total distance = > ut + (1/6)bt^3 m.

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The acceleration of a particle is increasing linearly with time t as bt. The particle starts from origin with initial velocity u. The distance travelled by the particle in time t will be: a) ut+(1/3)bt 2 b) ut+(1/2)bt 2 c) ut+(1/6)bt 3 d) ut+(1/3)bt 3

The acceleration of a particle is increasing linearly with time t as bt. The particle starts from origin with initial velocity u. The distance travelled by the particle in time t will be:

a) ut+(1/3)bt2

b) ut+(1/2)bt2

c) ut+(1/6)bt3

d) ut+(1/3)bt3

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The acceleration of a particle is increasing linearly with time t as bt. The particle starts from origin with initial velocity u. The distance travelled by the particle in time t will be: a) ut+(1/3)bt 2 b) ut+(1/2)bt 2 c) ut+(1/6)bt 3 d) ut+(1/3)bt 3

The acceleration of a particle is increasing linearly with time t as bt. The particle starts from origin with initial velocity u. The distance travelled by the particle in time t will be: a) ut+(1/3)bt2 b) ut+(1/2)bt2 c) ut+(1/6)bt3 d) ut+(1/3)bt3

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The acceleration of a particle is increasing linearly with time t as bt. The particle starts from origin with initial velocity u. The distance travelled by the particle in time t will be:

a) ut+(1/3)bt2

b) ut+(1/2)bt2

c) ut+(1/6)bt3

d) ut+(1/3)bt3