Explanation:
Given that,
Initial speed of the car, u = 88 km/h = 24.44 m/s
Reaction time, t = 2 s
Distance covered during this time,
(a) Acceleration,
We need to find the stopping distance, v = 0. It can be calculated using the third equation of motion as :
s = 74.66 meters
s = 74.66 + 48.88 = 123.54 meters
(b) Acceleration,
s = 37.33 meters
s = 37.33 + 48.88 = 86.21 meters
Hence, this is the required solution.
Answer:
Explanation:
Given ,
dv / dt = k ( 160 - v )
dv / ( 160 - v ) = kdt
ln ( 160 - v ) = kt + c , where c is a constant
when t = 0 , v = 0
Putting the values , we have
c = ln 160
ln ( 160 - v ) = kt + ln 160
ln ( 160 - v / 160 ) = kt
(160 - v ) / 160 =
1 - v / 160 =
v / 160 = 1 -
v = 160 ( 1 - )
differentiating ,
dv / dt = - 160k
acceleration a = - 160k
given when t = 0 , a = 280
280 = - 160 k
k = - 175
a = - 160 x - 175
a = 28000
when a = 128 t = ?
128 = 28000
= .00457