Question 365.
At time t = 0 a car has a velocity of 16 m/s. It slows down with an acceleration given by -0.50t, in m/s^2 for t in seconds. What is the distance travelled by the car by the time it stops?
Answer 365.
dv/dt = - 0.5 t
=> dv = - 0.5 t dt
Integrating,
v = - (1/4) t^2 + c
t = 0 => v = 16 => c = 16
=> v = - (1/4) t^2 + 16 ... ( 1 )
=> ds/dt = - (1/4) t^2 + 16
=> ds = [16 - (1/4) t^2] dt
Integrating,
s = 16t - (1/12) t^3 + c
t = 0 => s = 0 => c = 0
=> s = 16t - (1/12)t^3 ... ( 2 )
When it stops v = 0
=> from eqn. ( 1 ), 0 = - (1/4) t^2 + 16 => t = 8
Plugging this value of t in eqn. ( 2 ),
distance travelled, s
= 16 * 8 - (1/12) * 8*3
= 128 - 42.7
= 85.3 m.
Link to YA!
At time t = 0 a car has a velocity of 16 m/s. It slows down with an acceleration given by -0.50t, in m/s^2 for t in seconds. What is the distance travelled by the car by the time it stops?
Answer 365.
dv/dt = - 0.5 t
=> dv = - 0.5 t dt
Integrating,
v = - (1/4) t^2 + c
t = 0 => v = 16 => c = 16
=> v = - (1/4) t^2 + 16 ... ( 1 )
=> ds/dt = - (1/4) t^2 + 16
=> ds = [16 - (1/4) t^2] dt
Integrating,
s = 16t - (1/12) t^3 + c
t = 0 => s = 0 => c = 0
=> s = 16t - (1/12)t^3 ... ( 2 )
When it stops v = 0
=> from eqn. ( 1 ), 0 = - (1/4) t^2 + 16 => t = 8
Plugging this value of t in eqn. ( 2 ),
distance travelled, s
= 16 * 8 - (1/12) * 8*3
= 128 - 42.7
= 85.3 m.
Link to YA!
No comments:
Post a Comment