a motor car accelerate for 10sec to attain a velocity for 20msec it comtinue with uniform velocity for a further 20sec and then disccelerate so that it stop in 20sec,calculate.(1)acceleration,(2)deceleration
1. V = a * t = 20 m/s.
a * 10 = 20
a = 2 m/s^2.
2. V = Vo + a*t = 0
20 + a*20 = 0
20a = -20
a = -1.0 m/s^2.
To calculate the acceleration and deceleration, we need to apply the kinematic equations for motion. The relevant equation in this case is:
vf = vi + at
where vf is the final velocity, vi is the initial velocity, a is the acceleration or deceleration, and t is the time interval.
Given that the car accelerates for 10 seconds to attain a velocity of 20 m/s, we can use the equation to find the acceleration:
20 m/s = vi + a * 10 s
Since the car starts from rest (vi = 0), the equation simplifies to:
20 m/s = 0 + a * 10 s
20 m/s = 10 a
Dividing both sides by 10, we find:
a = 2 m/s²
So, the acceleration of the car is 2 m/s².
Next, we need to find the deceleration. We know that the car decelerates so that it comes to a stop in 20 seconds. We can use the same equation, but with the final velocity, vf, being 0 m/s:
0 m/s = vf + a * 20 s
Since vf = 20 m/s (from the previous information), the equation becomes:
0 m/s = 20 m/s + a * 20 s
Solving for a, we get:
a = -1 m/s²
Since the deceleration is in the opposite direction of motion, it is negative. So, the deceleration of the car is -1 m/s².