The displacement in metres of a particle on a straight line is s = 5t^3 + 3t + 8, with t in seconds. Find the velocity and the acceleration at t = 5 seconds.

Step-by-step solution

1. 1

   Recall what each derivative means. Velocity is the instantaneous rate of change of displacement, $v=\frac{ds}{dt}$, and acceleration is the rate of change of velocity, $a=\frac{dv}{dt}=\frac{d^2s}{dt^2}$. No kinematic formula is needed — differentiation supplies both.

2. 2

   Differentiate once for velocity. Applying the power rule term by term, $v(t)=15t^2+3$. The constant $8$ disappears because a fixed offset in position does not affect how fast the particle moves.

3. 3

   Differentiate again for acceleration. $a(t)=30t$. Since this is not constant, the motion is not uniformly accelerated.

4. 4

   Substitute $t=5$ into the velocity. $v(5)=15(5)^2+3=15(25)+3=375+3=378$ metres per second.

5. 5

   Substitute $t=5$ into the acceleration. $a(5)=30(5)=150$ metres per second squared.

6. 6

   Sanity-check the signs. Both values are positive, so at $t=5$ s the particle moves in the positive direction and is still speeding up, consistent with the strictly increasing cubic.