1. NAME: ALAKA PRECIOUS CHINONSO
COURSE: ENG102
DEPARTMENT: ENGINEERING (CIVIL)
LEMASS QUESTION
1. A tortoise and a hare are in a road race to defined the
honour of their breed. The tortoise craws the entire
1000m at a speed of 0.2m/s. the rabbit runs the first
200meters at 2m/s, stops to take a nap for 1.3hrs, and
awakens to finish the last 800meters with an average
speed of 3m/s. who wins the race and by how much
time?
Solution
Tortoise distance=1000m, speed=0.2m/s
Rabbit distance=200meters, speed=2m/s
Time for taking a nap=1.3*3600=4,680secs
After the nap distance covered=800meters,average speed
of 3m/s.
Tortoise= sav =distance/time.
0.2=1000/t
0.2t=1000
T=1000/2
Therefore t=5,000secs
Rabbit=sav=distance/time
2=200/t
2. 2t=200
Therefore t=100seconds
After nap time=1.3hrs=1.3*3600=4680
=4680+100=4780
sav of rabbit after nap, sav=distance/time
3=800/t
3t=800
T=800/3=266.67
T=266.67s
Total time taken by rabbit=266.67+4780
=5046.67secs
Total time= (5046.67-5000)
=46.67seconds.
Therefore the rabbit won the race at 46.67sec ahead of
rabbit.
2. If a particle’s position is given by x=6-12+14t2
(Where t is in secs and x is in meters)
1. What is the velocity at 1s
2. What is the speed at 1s
3. is there ever an instant when the velocity is o? if so
give the time.
Solution
X=6-12t+14t2
V=?
3. S=?
X=6-12*1+14(1)
X=6-12+14
X=-6+14
X=8
At t=1s, speed =av/t=8/1=8
Speed=8
X=6-12*1+14
X=6-12+14
X=-6+14
X=8
At t=1s velocity=8m/s
There is an instant when v=o
Therefore time=1s
3. The slowest animal ever discovered is a crab found in the
red sea that travels an average speed of 5.7km/year.
How long will it take this crab to travel 1meter?
Solution
sav =5.7km/year
Time to take the crab=?
Distance=1m
5.7km=5.7*1000=5700m
1yr=5700m
sav=distance/time
4. 5700=1/t
5700t=1
T=1/5700
T=0.000175seconds.
4. A car travels up a hill at a constant speed of 37km/h and
returns down the hill at a constant of 66km/h. calc. the
average speed for the whole trip.
Solution
V1=37km/hr
V2=66km/h
The first time,t1=d/2/v1
D /2 *1/v1
d/2*1/37=d/74
t2=d/2/v2
D /2 *1/v2
D /2 * 1/66=d/132
T2=d/132
Sum t1+t2
d/37+d/132
L.c.m of 37 and 132=4884
37d+66d/4884
103d/4884
Total time=0.0211d
5. sav =d/0.0211d
sav=2/0.211*2
sav=2/0.0422
sav=43.4km/hr
5. We drive a distance of 1km at 16km/h. then we drive an
additional distance of 1km at 32km/h. what is our
average speed
Solution
Distance of 1km at 16km/hour
Additional distance of 1km at 32km/hour
t1=d/2/t
t1=16/2*/t
t1=8/t
t2=d/2/t
t2=32/2*1/t
t2=16/t
sum 8/t + 16/t =16 + 8/t=24/t
sav =distance/time
sav=24/t
sav=24/2
sav=12km/hour
12*1000/3600=1200/3600
6. 3.333m/s
6. A car is initially travelling due north at 23m/s
A. find the velocity of the car after 4s if it’s acceleration is
2m/s2
due north.
B. find the velocity of the car after 4s if it’s acceleration is
instead 2m/s2
due south.
Solution
1. When u=23m/s
T=4s
A=2m/s2
V=?
Therefore we introduce the first equation of motion in this
case.
V=u + at
V=23 + 2(4)
V=23 + 8
V=31m/s due north.
2. V=u + at
31=u + 2(4)
31=u + 8
