Question

A particle moves in one dimension. Its velocity is given by \( v(t) = c_2 t^2 + c_1 t + c_0 \) where \( c_1 \) and \( c_2 \) are constants. What is the acceleration of the particle at time \( t = 1 \)?

• A. \( c_1 + 2 c_2 \)
• B. zero
• C. \( c_1 + c_2 \)
• D. \( c_1 \)

Hint:

To find the acceleration, differentiate the velocity equation with respect to time.

Answer 1

The Correct Option is A

The acceleration is the derivative of velocity with respect to time: \[ a(t) = \frac{dv(t)}{dt} \] Given \( v(t) = c_2 t^2 + c_1 t + c_0 \), we differentiate with respect to time: \[ a(t) = 2 c_2 t + c_1 \] At \( t = 1 \): \[ a(1) = 2 c_2 (1) + c_1 = c_1 + 2 c_2 \] Hence, the correct answer is (a).