Question

you drive with a constant speed of 13.5 m/s for 30 s. you then accelerate for 10s to a speed of 22 m/s you then slow to a stop in 10s how far have you traveled?

Answer 1

case I
when costt. v= 13.5 m/s
time taken,t = 30 s
dist. covered,\[s_{1}\]= v *t = 13.5 * 30 =405 m


case II
initial vel. , u = 13.5 m/s
final vel. , v= 22 m/s
time taken ,\[ t _{2}=10 s\]

thus acc. ,a = (v-u)/t = (22- 13.5)/10 = .085 m/s^2

hence, \[v ^{2}-u^{2}=2as\]
\[22^{2}-13.5^{2}= 2(0.85)*s\]
484-182.25 = 1.7 *s
\[s _{2}=177.5 m\]



caseIII
v=0
u=22 m/s
t=10 s
a=(v-u)/t = -2.2m/s^2
\[v ^{2}-u ^{2}=2as\]
-22 *22 /[2 *(-2.2)] =s

\[s_{3}=110 m\]


total dist.=

\[s _{1}+s _{2}+s _{3}= \]
405 + 177.5 + 110 = 692.5 m

Answer 2

the area under the graph gives the distance covered . this is 2nd method to calculate the distance traveled.
area = area(sq.) + ar(trapezium) + ar(triangle)
u get the 3 equations in a single step