Question

The velocity of a particle moving on the x axis is given by v=x^2+x where v is in m/s and x is in m find its acceleration in m/s^2 when passing through the point x=2m

Answer 1

Given:

• v = x² + x m/sec

To find :

• Acceleration at point x= 2 m

Solution:

• We know that, Acceleration (a) = d (Velocity) / d (time)
• a = d(v) / d(t)
• By applying chain rule,
• $\frac{dv}{dt} = \frac{dv}{dx} * \frac{dx}{dt}$    ...... (1)  
• d(v)/d(x) = d(x² + x) / d(x)
• d(v) / d(x) = 2 (x) + 1     .....(2)
• d(x) / d(t) is nothing but velocity which is given as x² + x  ..... (3)
• Substituting (2) and (3) in (1)
• $\frac{dv}{dt} = \frac{(2x + 1)}{1} * \frac{x^2 + x}{1}$  
• d(v) / d(t) = 2 x³+3 x²+ x
• ∴ Acceleration at any point = 2 x³ + 3 x² + x
• At point x = 2 m,
• a = 2 * (2)³ + 3 (2)² + 2
• a = 16 + 12 + 2
• a = 30 m / sec²

Answer:

The acceleration at point x = 2 m is 30 m/sec² .