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
Answer:
is 1
Step-by-step explanation:
sorry if im wrong this is my first time doing this
The rate of the water flowing in the channel was 2 miles per hour, the rate Evelyn and Meredith were paddling is mathematically given as
v+ = 2.29 1mile/hour
<h3>What is the rate Evelyn and
Meredith were paddling?</h3>
Generally, the equation for the up and downstream motion is mathematically given as
vup=v-2
vdown=v+2
Therefore, Statement interpretation
2/3=40*1/60
3+2/3hr=11/3
Where, tup+tdown=11/3
tup+tdown=11/3
1/Vup+1/vdown=11/3
Considering the LCD equation
3(v-2)(V+2)
Hence
11v^2-6v-44=0
Resolving using the equation we have
v+ = 2.29 1mile/hour
v- = -1.745 mile hour
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The area of the triangle is 144 m
2/4,6/12,1/4,3/12 make sure to label with miles cause it will probably be wrong hope this helps!
