The least common multiple is 32
Answer:
a) distance covered by hare d1 = 8t
b) distance covered by tortoise d2 = 5t + 550
c) ∆d = 550 - 3t
Step-by-step explanation:
Given;
Speed of hare u = 8m/s
Speed of tortoise v = 5 m/s
Initial distance of tortoise d0 = 550 m
a) using the equation of motion;
distance covered = speed × time + initial distance
d = vt + d0
For hare;
d0 = 0
Substituting the values;
d1 = 8t + 0
d1 = 8t
b)using the equation of motion;
distance covered = speed × time + initial distance
d2 = vt + d0
For tortoise;
d0 = 550m
Substituting the values;
d2 = 5t + 550
d2 = 5t + 550 m
c) the number of meters the tortoise is ahead of the hare.
∆d = distance covered by tortoise - distance covered by hare
∆d = d2 - d1
Substituting the values;
∆d = (5t + 550) - 8t
∆d = 550 - 3t
Could you please attach the graph itself because the question is not proper
Answer:
You didn't add a specific time frame so I can you a correct answer.
Explanation:
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Answer:
the answer is 60
Step-by-step explanation:
4*5*3=60
