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.

I think the answer is 360m is it correct experts

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

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

I gt it Mr. Damon

tnxx buddy

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.