A car starts from rest and moves along x-axis with constant acceleration (5m/secondsquare) fr 8 sec. If it then countinues with constant velocity. What distance will the car cover in 12 sec since it started frm rest. plz anybody explain me u gave me explaination before also bt it ws no enough plz tell me the answers also.
12 years ago
12 years ago
acceleration period:
v = a t = 5 * 8 = 40 m/s
x = (1/2) a t^2 = 2.5 * 64 = 160 m
this is the initial condition for the constant velocity coasting period:
x = Xi + Vi t + (1/2) a t^2
x = 160 + 40 t + (1/2) 0 t^2
= 160 + 40(12-8)
= 160 + 160
= 320 meters
12 years ago
tnx Mr. Damon
bt i don't think the explaination is fr a class 9th student cn it be mr easy?? as i don't know what is Xi and Vit
12 years ago
I gt it Mr. Damon
tnxx buddy
11 months ago
To find the distance the car will cover in 12 seconds since it started from rest, you need to break down the problem into two parts: the time interval when the car is accelerating and the time interval when the car is moving at constant velocity.
1. When the car is accelerating (initial 8 seconds):
To find the distance covered during this acceleration phase, you can use the equation of motion:
s = ut + (1/2)at^2
Where:
s = distance covered
u = initial velocity (0 m/s, since the car starts from rest)
t = time (8 seconds)
a = acceleration (5 m/s^2)
Plugging in the values, we get:
s = 0 + (1/2)(5)(8)^2
s = 0 + (1/2)(5)(64)
s = 0 + 5(32)
s = 160 meters
Therefore, during the acceleration phase, the car covers a distance of 160 meters.
2. When the car is moving at constant velocity (last 4 seconds):
Since the car is moving at constant velocity, the distance covered can be found using the formula:
s = vt
Where:
s = distance covered
v = velocity (which remains constant after acceleration phase)
t = time (4 seconds)
To find the constant velocity, we can use the formula:
v = u + at
Where:
v = final velocity (which is constant after acceleration phase)
u = initial velocity (which is 0 m/s, since the car starts from rest)
a = acceleration (given as 5 m/s^2)
t = time of acceleration (8 seconds)
Plugging in the values, we get:
v = 0 + (5)(8)
v = 0 + 40
v = 40 m/s
Now we can calculate the distance covered during constant velocity:
s = (40)(4)
s = 160 meters
Therefore, during the constant velocity phase, the car covers a distance of 160 meters.
3. Total distance covered:
To find the total distance covered, we add up the distances from the acceleration phase and the constant velocity phase:
Total distance = Distance during acceleration + Distance during constant velocity
Total distance = 160 meters + 160 meters
Total distance = 320 meters
Therefore, in 12 seconds since it started from rest, the car will cover a total distance of 320 meters.