Engineering Mechanics - Kinematics of Particle (KOP) - Discussion
Discussion Forum : Kinematics of Particle (KOP) - General Questions (Q.No. 1)
1.
A race car starting from rest moves along a straight track with an acceleration as shown in the graph (where for t 
10 s, a = 8 m/s2). Determine the time t for the car to reach a speed of 50 m/s.
Discussion:
16 comments Page 2 of 2.
Santhosha said:
1 decade ago
Initially the is at rest, we consider u=40m/s, and v=50m/s then
v=u+at
50=40+8*t
t=1.25sec,
This is the time required to travel from 40m/s velocity to 50m/s velocity. then the time required to travel the car from rest position is 10+1.25=11.25 sec.
v=u+at
50=40+8*t
t=1.25sec,
This is the time required to travel from 40m/s velocity to 50m/s velocity. then the time required to travel the car from rest position is 10+1.25=11.25 sec.
Mani said:
1 decade ago
S=1/2 (8+0) 10.
Since initial velocity is zero. My guess.
Since initial velocity is zero. My guess.
Gopala krishna said:
1 decade ago
Here the car is not moving with uniform accleration upto ten seconds
area of the triangle=0.5*8*10=40m/sec(t=10)
now accleration is constant
So apply v=u+at
50=40+8*t
t=1.25
So total timet = 10+1.25 = 11.25
area of the triangle=0.5*8*10=40m/sec(t=10)
now accleration is constant
So apply v=u+at
50=40+8*t
t=1.25
So total timet = 10+1.25 = 11.25
Puru said:
1 decade ago
Dear, please tell me how it comes 40m/s. By which law and logic you calculate 40m/s.
FYI
1. V=u+at
3. v2=u2+2as
2. s=ut+1/2at2
4. s=1/2(v+u)t
FYI
1. V=u+at
3. v2=u2+2as
2. s=ut+1/2at2
4. s=1/2(v+u)t
Abhishek chouksey said:
1 decade ago
Its Simple>
in first 10 sec it reaches only 40 m/s=1/2*10*8(uniform acc)
in next 1.25 sec at constant acc.(v=u+at) it tends to 50 m/s.
in first 10 sec it reaches only 40 m/s=1/2*10*8(uniform acc)
in next 1.25 sec at constant acc.(v=u+at) it tends to 50 m/s.
Veni said:
1 decade ago
How answer will come? Please explain.
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