Civil Engineering - Applied Mechanics - Discussion
Discussion Forum : Applied Mechanics - Section 2 (Q.No. 44)
44.
Time required to stop a car moving with a velocity 20 m/sec within a distance of 40 m, is
Discussion:
9 comments Page 1 of 1.
Engr Muhammad Khalid said:
1 year ago
Solution:
Data given:
V = 20 m/sec2
S = 40m.
Required
Time, T = ?.
Solution:
By using Formula;
S=V×T ( V = v1+v2/2)
S= 10T (V = 20+0/2)
40/10= T (V = 10 m/Sec).
T = 4 Sec Ans.
Data given:
V = 20 m/sec2
S = 40m.
Required
Time, T = ?.
Solution:
By using Formula;
S=V×T ( V = v1+v2/2)
S= 10T (V = 20+0/2)
40/10= T (V = 10 m/Sec).
T = 4 Sec Ans.
Mian Janealam said:
4 years ago
Thanks @Asay.
Shyamlal said:
5 years ago
S = (u+v) t/2.
Take u = 0,
Solve it t = 4 sec.
Take u = 0,
Solve it t = 4 sec.
(2)
Syed masroor ali said:
6 years ago
Simply s=vt v=v1+v2/2 = 10m/s.
So
40 = 10t.
t = 4s.
So
40 = 10t.
t = 4s.
(1)
Vishal kamboj said:
6 years ago
Using the third equation we will find acceleration.
{v }^{2} = {u}^{2} + 2as \\ 0 = 400 - 80a.
So a = 5.
Now using the first equation we will find the time,
v=u+at,
0=20-5t,
so t=4 seconds.
{v }^{2} = {u}^{2} + 2as \\ 0 = 400 - 80a.
So a = 5.
Now using the first equation we will find the time,
v=u+at,
0=20-5t,
so t=4 seconds.
Harry said:
6 years ago
Thanks alot @Asay.
Shay said:
7 years ago
Thank you @Asay.
Epcet Amit said:
7 years ago
Nice, Thanks @Asay.
Asay said:
9 years ago
Given Data:
vi = 20m/sec
vf = 0m/sec
S = 40m
t = ?
Solution:
We know that,
S=vi * t + 1/2at^2.
But , Here , a=?
We know that,
2aS = vf^2 - vi^2.
2 * a * 40 = 0 - 20^2.
a = - 400/80 = -5m/sec^2.
Now putting values in eqn S = vi * t + 1/2at^2.
40m = 20 * t + (-5 * t^2)/2.
40m = 20t - 2.5t^2.
25t^2 - 200t + 400 = 0.
t^2 - 8t + 16 = 0.
(t-4)^2=0.
t - 4 = 0.
t = 4sec-> Ans
vi = 20m/sec
vf = 0m/sec
S = 40m
t = ?
Solution:
We know that,
S=vi * t + 1/2at^2.
But , Here , a=?
We know that,
2aS = vf^2 - vi^2.
2 * a * 40 = 0 - 20^2.
a = - 400/80 = -5m/sec^2.
Now putting values in eqn S = vi * t + 1/2at^2.
40m = 20 * t + (-5 * t^2)/2.
40m = 20t - 2.5t^2.
25t^2 - 200t + 400 = 0.
t^2 - 8t + 16 = 0.
(t-4)^2=0.
t - 4 = 0.
t = 4sec-> Ans
(2)
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