1.State the three motion equations?
Answer:
V=u+at....1
S =ut +.5at²
V²=u²2as
2.What do we mean by the initial velocity and final velocity?
Answer:
Initial velocity is the velocity which the body has in the beginning of the given time period and final velocity is the velocity which the body has at the end of the given time period.
3. A car starts from rest and accelerates uniformly over a time of 5.21 seconds for a A distance of 110 m. Determine the acceleration of the car.
Answer:
given,
u=0
t=5.21
s=110m
we know s=1/2at²
110=1/2 a(5.21)²
220/5.21²=a
a=8.104m/s²
acceleration of the car is 8.1 m/s²
4.A car starts from rest and accelerates at 9.54m/s? for 6.5 seconds. What is the distance covered by the car?
Answer:
S =ut + 1/2at²
S = 0×6.5 + (1/2 × 9.54) × 6.5²
S =0 + 4.77 ×42.25
S=201.5m
5.An eagle flies at a velocity of 40m/s toward mark district; it accelerates at 15m/s? Calculate the final velocity after 10 seconds.
Answer
Given:
u=40m/s a=15m/s² t=10s V=?
V = u+at
V= 40m/s+15m/s²×10s
V=40m/s+150m/s
V=190m/so
6. A racing car accelerates uniformly from 18.5 m/s to 46.1 m/s in 2.47 seconds. Determine the acceleration of the car and the distance traveled.
Answer
Given u=18.5m/s v=46.1m/s t= 2.47s a=?
a= v – u÷t
46.1m/s-18.5m/s÷2.47s
a=27.6m/s÷2.47s
=11.174m/s²
Now, a = v/t
= 46.1 - 18.5 / 2.47
a = 11.2 m/s²
S = u× t + 0.5 × a × t²
S= 18.5 m/s × 2.47 + 0.5 × (11.2 m/s²) × (2.47 s)²
S= 45.7 m + 34.1 m
S = 79.8 m
7. An airplane accelerates down a runway at 3.20m/s² for 32.8 s until it finally lifts off the ground. Determine the distance traveled before take-off.
Answer
Given: u=0m/s a=3.20m/s² t=32.8s D=?
S=ut+0.5×at²
s = 0.5at²
s=1/2(3.20)(32.8)²
s= 1721.344 m
8. Derive the following equation
a) v=u+at
Answer
b) v²=u²+2as
Answer
9.A snail covers a distance of 100 metres in 50 hours. Calculate the average speed of snail in km/h.
Answer
Given: d=100m t=50html
Distance÷time
100m÷50hr
=2m/h
2÷100
=0.002km/h
10.If a sprinter runs a distance of 100 metres in 9.83 seconds, calculate his average speed in km/h.
Answer
Given: D=100m t=9.83
=average speed?
=100m÷9.83s
=10.17m/s
36.622km/h
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