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A particle p is moving on a straight line. The end O is the fixed point on the line. The displacement of p from O at time(t) seconds is (S) metres, where S = 6 + 10t − t2. Find a) the distance travelled from t=1 to t=4 b) the velocity when t=6

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Answers

2 parts
a

15 m\text{15 m}

b

-2 m/s\text{-2 m/s}

Step-by-step solution

6 steps

The displacement of the particle is given by S=6+10t−t2S = 6 + 10t - t^2.

Part a· 4 steps

The distance travelled from t=1t=1 to t=4t=4

  1. Find the velocity function by differentiating the displacement function with respect to time.

    V=dSdt=ddt(6+10t−t2)=10−2tV = \frac{dS}{dt} = \frac{d}{dt}(6 + 10t - t^2) = 10 - 2t

  2. Determine if the particle changes direction within the interval t=1t=1 to t=4t=4 by finding when the velocity is zero.

    Set V=0V = 0:

    10−2t=010 - 2t = 0 2t=102t = 10 t=5 st = 5 \text{ s}

    Since t=5t=5 is outside the interval [1,4][1, 4], the particle does not change direction during this time. Therefore, the distance travelled is the magnitude of the displacement between t=1t=1 and t=4t=4.

  3. Calculate the displacement at t=1t=1 and t=4t=4.

    At t=1t=1:

    S(1)=6+10(1)−(1)2=6+10−1=15 mS(1) = 6 + 10(1) - (1)^2 = 6 + 10 - 1 = 15 \text{ m}

    At t=4t=4:

    S(4)=6+10(4)−(4)2=6+40−16=30 mS(4) = 6 + 10(4) - (4)^2 = 6 + 40 - 16 = 30 \text{ m}

  4. Calculate the distance travelled.

    Distance travelled =∣S(4)−S(1)∣=∣30−15∣=15 m= |S(4) - S(1)| = |30 - 15| = 15 \text{ m}
    The distance travelled from t=1t=1 to t=4t=4 is 15 m\boxed{\text{15 m}}.

Part b

-2 m/s\text{-2 m/s}

The velocity when t=6t=6

  1. Use the velocity function found in part a).

    V=10−2tV = 10 - 2t

  2. Substitute t=6t=6 into the velocity function.

    V(6)=10−2(6)=10−12=−2 m/sV(6) = 10 - 2(6) = 10 - 12 = -2 \text{ m/s}

    The velocity when t=6t=6 is -2 m/s\boxed{\text{-2 m/s}}.

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