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Exercise 7.1 · Q3

Q.A particle moves along a line according to the law s(t)=2t3−9t2+12t−4s(t)=2t^3-9t^2+12t-4, where t≥0t\ge0.

(i) At what times the particle changes direction?
(ii) Find the total distance travelled by the particle in the first 4 seconds.
(iii) Find the particle's acceleration each time the velocity is zero.
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v(t)=s′(t)v(t)=s'(t) tells us when and where the particle changes direction (sign changes of vv); the total distance adds the absolute value of each directional "leg"; the acceleration a(t)=v′(t)a(t)=v'(t) is evaluated at each time v=0v=0.

Step 1. Differentiate for velocity and acceleration.

s(t)=2t3−9t2+12t−4⇒v(t)=6t2−18t+12=6(t2−3t+2)=6(t−1)(t−2)s(t)=2t^3-9t^2+12t-4\Rightarrow v(t)=6t^2-18t+12=6(t^2-3t+2)=6(t-1)(t-2), and a(t)=12t−18a(t)=12t-18.

Step 2. Find when the particle changes direction.

v(t)=0v(t)=0 at t=1,2t=1,2. For t<1t<1: both factors negative ⇒v>0\Rightarrow v>0. For 1<t<21<t<2: (t−1)>0,(t−2)<0⇒v<0(t-1)>0,(t-2)<0\Rightarrow v<0. For t>2t>2: both positive ⇒v>0\Rightarrow v>0. Since vv genuinely changes sign at both roots, the particle changes direction at t=1t=1 and t=2t=2.

Step 3. Total distance in [0,4][0,4].

Compute ss at 0,1,2,40,1,2,4: s(0)=−4s(0)=-4, s(1)=2−9+12−4=1s(1)=2-9+12-4=1, s(2)=16−36+24−4=0s(2)=16-36+24-4=0, s(4)=128−144+48−4=28s(4)=128-144+48-4=28.

Total distance=∣s(1)−s(0)∣+∣s(2)−s(1)∣+∣s(4)−s(2)∣=∣1−(−4)∣+∣0−1∣+∣28−0∣=5+1+28=34 m.\text{Total distance}=|s(1)-s(0)|+|s(2)-s(1)|+|s(4)-s(2)|=|1-(-4)|+|0-1|+|28-0|=5+1+28=34\text{ m}.

Step 4. Acceleration at each time the velocity is zero.

a(1)=12(1)−18=−6a(1)=12(1)-18=-6; a(2)=12(2)−18=6a(2)=12(2)-18=6.

✓Final answer

  1. The particle changes direction at t=1t=1 s and t=2t=2 s.
  2. Total distance travelled in the first 4 seconds is 3434 m.
  3. Acceleration is −6-6 m/s² at t=1t=1 and 66 m/s² at t=2t=2.

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