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The acceleration of a particle increases linearly with time t as 6t. If the initial velocity of the particle is zero and the particle starts from the origin, then the distance travelled by the particle in time t will be
- a)t
- b)t2
- c)t3
- d)t4
Correct answer is option 'C'. Can you explain this answer?
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The acceleration of a particle increases linearly with timet as6t. If ...
Explanation:
Given:
- Acceleration of the particle: a(t) = 6t
- Initial velocity of the particle: 0
- Particle starts from the origin
Using kinematic equations:
- We know that acceleration is the derivative of velocity with respect to time, which means:
a(t) = dv/dt
- Integrating both sides with respect to time, we get:
∫a(t) dt = ∫dv
- Integrating the given acceleration function, we get:
∫6t dt = v
v = 3t^2
- Since the initial velocity is 0, we have v(0) = 0, which means the velocity function is:
v(t) = 3t^2
- Distance travelled by the particle is the integral of velocity with respect to time:
∫v(t) dt = ∫3t^2 dt
∫3t^2 dt = t^3
Therefore, the distance travelled by the particle in time t is t^3, which corresponds to option 'c'.
Given:
- Acceleration of the particle: a(t) = 6t
- Initial velocity of the particle: 0
- Particle starts from the origin
Using kinematic equations:
- We know that acceleration is the derivative of velocity with respect to time, which means:
a(t) = dv/dt
- Integrating both sides with respect to time, we get:
∫a(t) dt = ∫dv
- Integrating the given acceleration function, we get:
∫6t dt = v
v = 3t^2
- Since the initial velocity is 0, we have v(0) = 0, which means the velocity function is:
v(t) = 3t^2
- Distance travelled by the particle is the integral of velocity with respect to time:
∫v(t) dt = ∫3t^2 dt
∫3t^2 dt = t^3
Therefore, the distance travelled by the particle in time t is t^3, which corresponds to option 'c'.
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