a car runs with a negative acceleration of -0.5m/s^2. the initial velocity of the car is 54 km/hr. In how much time and how far from the initial point will the car stop?
Answers
Answered by
11
From question it is clear that
acceleration (a)= - 0.5 m/s²
initial velocity (u) = 54 km/hr
As the acceleration is given in m/s² let us convert the initial velocity also into m/s²
therefore
initial velocity = 15 m/s
1)We have to find the time(t) taken by the car to get stopped
From v = u + at
the final velocity(v) = 0 . since the car is retar ding
0 = 15 + (- 0.5) t
0.5*t = 15
t = 30 seconds. This is time taken by the car to stop.
2)Now we find the displacement(s) covered by the car before it stops
From v² - u² = 2as
by substituting we get
0 - (15)² = 2(- 0.5)s
1*s = 225
s = 225 m . This the displacement covered by car before it stops .
acceleration (a)= - 0.5 m/s²
initial velocity (u) = 54 km/hr
As the acceleration is given in m/s² let us convert the initial velocity also into m/s²
therefore
initial velocity = 15 m/s
1)We have to find the time(t) taken by the car to get stopped
From v = u + at
the final velocity(v) = 0 . since the car is retar ding
0 = 15 + (- 0.5) t
0.5*t = 15
t = 30 seconds. This is time taken by the car to stop.
2)Now we find the displacement(s) covered by the car before it stops
From v² - u² = 2as
by substituting we get
0 - (15)² = 2(- 0.5)s
1*s = 225
s = 225 m . This the displacement covered by car before it stops .
Answered by
1
Explanation:
answer is 30 second and 225 metre
Similar questions