Velocity and acceleration questions with answers

Questions

1. Question 1Velocity from displacement

   A particle moves in a straight line so that, $t$ seconds after passing through a fixed point $O$, its displacement from $O$, $s$ m, is given by $s = -t^{3} + 18t^{2} - 103t$. Find its velocity when $t = 6$.

   Show the worked solution

   1. Step 1: Find v from s: differentiate or integrate?$v = \dfrac{ds}{dt}$
   2. Step 2: Find v$v = -3t^{2} + 18(2)t - 103(1)$$= -3t^{2} + 36t - 103$
   3. Step 3: Put the given t into v$v = -3(6)^{2} + 36(6) - 103$$= 5$ m/s
2. Question 2Velocity from displacement

   A particle moves in a straight line so that, $t$ seconds after passing through a fixed point $O$, its displacement from $O$, $s$ m, is given by $s = -t^{3} + 15t^{2} - 72t$. Find its velocity when $t = 7$.

   Show the worked solution

   1. Step 1: Find v from s: differentiate or integrate?$v = \dfrac{ds}{dt}$
   2. Step 2: Find v$v = -3t^{2} + 15(2)t - 72(1)$$= -3t^{2} + 30t - 72$
   3. Step 3: Put the given t into v$v = -3(7)^{2} + 30(7) - 72$$= -9$ m/s
3. Question 3Acceleration

   A particle moves in a straight line so that, $t$ seconds after passing through a fixed point $O$, its displacement from $O$, $s$ m, is given by $s = t^{3} - 18t^{2} + 59t$. Find its acceleration when $t = 2$.

   Show the worked solution

   1. Step 1: Find v from s: differentiate or integrate?$v = \dfrac{ds}{dt}$
   2. Step 2: Find v$v = 3t^{2} - 18(2)t + 59(1)$$= 3t^{2} - 36t + 59$
   3. Step 3: Find a from v: differentiate or integrate?$a = \dfrac{dv}{dt}$
   4. Step 4: Find a$a = 3(2)t - 36(1)$$= 6t - 36$
   5. Step 5: Put the given t into a$a = 6(2) - 36$$= -24$ m/s²
4. Question 4Acceleration

   A particle moves in a straight line so that, $t$ seconds after passing through a fixed point $O$, its displacement from $O$, $s$ m, is given by $s = t^{3} - 12t^{2} + 36t$. Find its acceleration when $t = 2$.

   Show the worked solution

   1. Step 1: Find v from s: differentiate or integrate?$v = \dfrac{ds}{dt}$
   2. Step 2: Find v$v = 3t^{2} - 12(2)t + 36(1)$$= 3t^{2} - 24t + 36$
   3. Step 3: Find a from v: differentiate or integrate?$a = \dfrac{dv}{dt}$
   4. Step 4: Find a$a = 3(2)t - 24(1)$$= 6t - 24$
   5. Step 5: Put the given t into a$a = 6(2) - 24$$= -12$ m/s²
5. Question 5Instantaneously at rest

   A particle moves in a straight line so that, $t$ seconds after passing through a fixed point $O$, its displacement from $O$, $s$ m, is given by $s = t^{3} - 6t^{2} + 9t$. Find the values of $t$ at which the particle is instantaneously at rest.

   Show the worked solution

   1. Step 1: Find v from s: differentiate or integrate?$v = \dfrac{ds}{dt}$
   2. Step 2: Find v$v = 3t^{2} - 6(2)t + 9(1)$$= 3t^{2} - 12t + 9$
   3. Step 3: Instantaneously at rest: which quantity is 0?At rest: $v = 0$.
   4. Step 4: Set it to 0. Simplify and factorise$3t^{2} - 12t + 9 = 0$$t^{2} - 4t + 3 = 0$$(t - 1)(t - 3) = 0$$t = 1$ or $t = 3$
6. Question 6Instantaneously at rest

   A particle moves in a straight line so that, $t$ seconds after passing through a fixed point $O$, its displacement from $O$, $s$ m, is given by $s = 2t^{3} - 30t^{2} + 54t$. Find the values of $t$ at which the particle is instantaneously at rest.

   Show the worked solution

   1. Step 1: Find v from s: differentiate or integrate?$v = \dfrac{ds}{dt}$
   2. Step 2: Find v$v = 2(3)t^{2} - 30(2)t + 54(1)$$= 6t^{2} - 60t + 54$
   3. Step 3: Instantaneously at rest: which quantity is 0?At rest: $v = 0$.
   4. Step 4: Set it to 0. Simplify and factorise$6t^{2} - 60t + 54 = 0$$t^{2} - 10t + 9 = 0$$(t - 1)(t - 9) = 0$$t = 1$ or $t = 9$