ThIs is the same type of problem
find out the time value
3 = 1/2*a*T^2
6/10 = t^2
t = 0.77 seconds
and the distance is given 5 m
thus speed ,= distance/time
speed = 5/0.77
= 6.45 m/s
Answer:
a) t = 4.16 s
b) x = 141.51 m
Explanation:
Given
v = 21.5 m/s
x0 = 52.0 m
a = 6.0 m/s²
a) Motorcycle
x = v0*t + (a*t²/2)
x = 21.5t + (6*t²/2)
x = 21.5t + 3t² <em>(I)</em>
Car
x = x0 + v0*t
x = 52 + 21.5t <em>(II)</em>
<em />
then we can apply <em>I = II</em>
21.5t + 3t² = 52 + 21.5t
⇒ 3t² = 52
⇒ t = 4.16 s
b) We can use <em>I</em> or <em>II</em>, then
x = 52 + 21.5*(4.16)
⇒ x = 141.51 m
<h2>
Its velocity when it crosses the finish line is 117.65 m/s</h2>
Explanation:
We have equation of motion s = ut + 0.5 at²
Initial velocity, u = 0 m/s
Acceleration, a = ?
Time, t = 6.8 s
Displacement, s = 1/4 mi = 400 meters
Substituting
s = ut + 0.5 at²
400 = 0 x 6.8 + 0.5 x a x 6.8²
a = 17.30 m/s²
Now we have equation of motion v = u + at
Initial velocity, u = 0 m/s
Final velocity, v = ?
Time, t = 6.8 s
Acceleration, a = 17.30 m/s²
Substituting
v = u + at
v = 0 + 17.30 x 6.8
v = 117.65 m/s
Its velocity when it crosses the finish line is 117.65 m/s
