Question

A particle moving in a straight line with uniform deceleration has a velocity of 40m/s at a point P, 20m/s at a point Q and comes to rest at a point R where QR=50m. Calculate the distance PQ, calculate the time taken to cover PQ and the time taken to cover PR

Answer 1

A particle moving in a straight line with uniform deceleration has a velocity of $40\mathrm{m/s}$ at a point P, $20\mathrm{m/s}$ at a point Q and comes to rest at a point R where QR=50m. Calculate the distance PQ, calculate the time taken to cover PQ and the time taken to cover PR

Solution:

The equation of motion along the X-axis for the distance QR:

$$x: Q R = V _ {Q} t _ {Q R} - \frac {a t _ {Q R} {} ^ {2}}{2}$$

Rate equation along the X-axis for the distance QR:

$$x: 0 = V _ {Q} - a t _ {Q R}$$ $$t _ {Q R} = \frac {V _ {Q}}{a} (2)$$ $$(2) \text{in} (1): Q R = V _ {Q} \frac {V _ {Q}}{a} - \frac {a \left(\frac {V _ {Q}}{a}\right) ^ {2}}{2}$$ $$Q R = \frac {V _ {Q} {} ^ {2}}{a} - \frac {V _ {Q} {} ^ {2}}{2 a}$$ $$Q R = \frac {V _ {Q} {} ^ {2}}{2 a}$$ $$a = \frac {V _ {Q} {} ^ {2}}{2 \cdot Q R} = \frac {\left(2 0 \frac {m}{s}\right) ^ {2}}{2 \cdot 5 0 m} = 4 \frac {m}{s ^ {2}}$$

We have found deceleration, you can now find the time $t_{QR}$, $t_{PQ}$ and distance PQ.

$$(2): t _ {Q R} = \frac {V _ {Q}}{a} = \frac {2 0 \frac {m}{s}}{4 \frac {m}{s ^ {2}}} = 5 s$$

The equation of motion along the X-axis for the distance PQ:

$$x: P Q = V _ {P} t _ {P Q} - \frac {a t _ {P Q} {} ^ {2}}{2}$$

Rate equation along the X-axis for the distance QR:

$$x: V _ {Q} = V _ {P} - a t _ {P Q}$$ $$t _ {P Q} = \frac {V _ {P} - V _ {Q}}{a} = \frac {4 0 \frac {m}{s} - 2 0 \frac {m}{s}}{4 \frac {m}{s ^ {2}}} = 5 s (4)$$ $$(4) i n (3): P Q = V _ {P} t _ {P Q} - \frac {a t _ {P Q} {} ^ {2}}{2} = 4 0 \frac {m}{s} \cdot 5 s - \frac {4 \frac {m}{s ^ {2}} \cdot (5 s) ^ {2}}{2} = 1 5 0 m$$

Answer: $t_{QR} = 5s$

$$t _ {P Q} = 5 s$$ $$P Q = 1 5 0 m$$