Kinematics of Motion in a Straight Line: Question 3

Worked solution

Part (a): Integrating acceleration to find velocity

Since $a = \dfrac{dv}{dt}$, velocity is found by integrating acceleration with respect to $t$:

$v = \int (6-3t)\,dt = 6t - \frac{3}{2}t^2 + C = 6t - 1.5t^2 + C$

The robot starts from rest at $O$, so $v=0$ when $t=0$:

$0 = 6(0) - 1.5(0)^2 + C \implies C = 0$

Therefore:

$v = 6t - 1.5t^2$

as required.

Recompute independently as a check: differentiating $v=6t-1.5t^2$ back gives $\dfrac{dv}{dt} = 6 - 3t$, which matches the given $a$ exactly, and $v(0)=6(0)-1.5(0)^2=0$ matches the “starts from rest” condition. Both checks confirm the result.

Part (b): Maximum velocity

The velocity is increasing while the acceleration is positive, and decreasing once the acceleration becomes negative. So the maximum velocity occurs at the instant the acceleration is zero (the turning point of $v$):

$a = 6 - 3t = 0 \implies t = 2$

Substituting $t=2$ into the velocity expression from part (a):

$v = 6(2) - 1.5(2)^2 = 12 - 1.5(4) = 12 - 6 = 6$

So the maximum velocity is $6\text{ m s}^{-1}$, occurring at $t=2$ s.

Justifying it is a maximum: $\dfrac{d^2v}{dt^2} = \dfrac{da}{dt} = -3 < 0$, so $t=2$ is a maximum turning point of $v$ (not a minimum). Equivalently, the acceleration $a=6-3t$ is positive for $t<2$ (velocity still increasing) and negative for $t>2$ (velocity decreasing), confirming $t=2$ gives the greatest velocity.

Part (c): Total distance travelled

Since $v = \dfrac{ds}{dt}$, the displacement is found by integrating the velocity:

$s = \int (6t - 1.5t^2)\,dt = 3t^2 - 0.5t^3 + C'$

The robot starts at $O$, so $s=0$ when $t=0$, giving $C'=0$:

$s = 3t^2 - 0.5t^3$

At $t=4$:

$s(4) = 3(4)^2 - 0.5(4)^3 = 3(16) - 0.5(64) = 48 - 32 = 16$

Before quoting this as the distance travelled, check that the robot never reverses direction on $0\le t\le 4$: from part (a), $v=6t-1.5t^2 = 1.5t(4-t)$, which is zero only at $t=0$ and $t=4$ and positive in between (it is a “cap”-shaped quadratic in $t$). So $v\ge 0$ throughout, meaning the robot always moves the same way. The distance travelled equals the displacement.

$\text{distance travelled} = s(4) - s(0) = 16 - 0 = 16\text{ m}$

Final answers

• (a) $v = \boxed{6t - 1.5t^2}$
• (b) Maximum velocity $=\boxed{6\text{ m s}^{-1}}$ at $t=\boxed{2}$ s
• (c) Total distance travelled $=\boxed{16\text{ m}}$