Respuesta :
Answer:
12.5 s , 312.5 m
Explanation:
initial velocity = u= 50 m/s
final velocity= v = 0
acceleration = a - -4 m/s^2
time= t=?
distance = s=?
equation of motion is v = u + at
0r at = v-u
t = (v-u)/a
t= (0-50)/-4
t= -50/-4
t= 12.5 s
s= ut +(1/2)at^2
s= 50(12.5)+(0.5)(-4)(12.5)^2
s=625-312-5
s =312.5 m
Answer:
Time taken to stop is 12.5 seconds.
Distance travelled is 312.5 m.
Explanation:
Initial velocity of the car = u = 50 m/s.
Acceleration of the car = a = -4 m/s² (negative because it decelerates)
Final velocity = v = 0 m/s (as the car stops)
To find:
- time t taken to stop
- distance s travelled till it stops.
By equations of motion:
v = u + at
s = ut+[tex]\frac{1}{2}[/tex]at²
t = [tex]\frac{v-u}{a}=\frac{0-50}{-4}=12.5\ s[/tex]
Hence time taken to stop is 12.5 seconds.
[tex]s=ut+\frac{1}{2}at^{2}=(50\times12.5)+\frac{1}{2}\times-4\times(12.5)^{2}=312.5\ m[/tex]
Distance travelled is 312.5 m.
