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statuscvo [17]
3 years ago
9

One speed skater starts across a frozen lake at an average speed of 8 m/s. Ten seconds later, a second speed skater starts from

the same point and skates in the same direction at an average speed of 10 m/s. How many seconds after the second skater starts will the second skater overtake the first skater?
Physics
1 answer:
Luba_88 [7]3 years ago
7 0

Answer:

40 s

Explanation:

After 10 seconds, the first skater would have a 8m/s * 10s = 80 m head start

Let t be the number of seconds after the second skater starts will the second skater overtake the first skater

The distance traveled by the first skater after t seconds is

s_1 = v_1t = 8t

Similarly the distance traveled by the 2nd skater after t seconds is

s_2 = v_2t = 10t

Since the 2nd skater catches up to the 1st one after 80 m behind, the distance traveled by the 2nd one must be 80m greater than the distance of the 1st skater

s_2 = s_1 + 80

We can substitute s_1 = 8t, s_2 = 10t

10t = 8t + 80

2t = 80

t = 80 / 2 = 40 s

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Answer:

(a) 2.5 m/s

(b) 37.5 KJ

Explanation:

(a)

From the law of conservation of momentum, Initial momentum=Final momentum

mV_1+3mV_2=(m+3m)V_f=4mV_f

V_1+3V_2=4V_f and making V_f the subject then

V_f=\frac {V_1+3V_2}{4} and since V_1 is initial velocity of car, value given as 4 m/s, V_2 is the initial velocity of the three cars stuck together, value given as 2 m/s and v_f is the final velocity which is unknown. By substitution

V_f=\frac {4+(3\times2)}{4}=2.5 m/s

(b)

Initial kinetic energy is given by

\frac {mV_1^{2}}{2}+\frac {3mV_2^{2}}{2}=\frac {m(V_1^{2}+3V_2^{2}}{2}=\frac {2.5\times 10^{4}(4^{2}+3(2^{2}))}{2}=350\times10^{3} J= 350 KJ

Final kinetic energy is given by

\frac {4mV_f^{2}}{2}=\frac {4\times 2.5\times 10^{4}\times 2.5^{2}}{2}=312.5\times 10^{3} J=312.5 KJ

The energy lost is given by subtracting the final kinetic energy from the initial kinetic energy hence

Energy lost=350-312.5=37.5 KJ

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