a car traveling on a straight road accelerates uniformly at a meter per second square from its initial speed of 20 meter per second until it reaches a speed of 30 meter per second. the car travels at this speed for 600 meter and then decelerates uniformly at 0.5a meter per second square until it comes to rest.show that the time spent decelerating is six times as long as the time spent accelerating. the total distance traveled by the car from time t=0 until it comes to rest in 2900 meter. find the time for which the car is accelerating.find the value of a. find the speed of the car 40 second before it comes to rest.
To find the time spent accelerating, we can use the equation of uniformly accelerated motion:
v = u + at
where:
v = final velocity = 30 m/s
u = initial velocity = 20 m/s
a = acceleration in m/s^2
Rearranging the equation to solve for time (t):
t = (v - u) / a
Substituting the given values:
t = (30 - 20) / a
t = 10 / a
To find the time spent decelerating, we can use the same equation:
t' = (v - u) / (-0.5a)
where:
v = final velocity = 0 m/s
u = initial velocity = 30 m/s
a = acceleration = 0.5a (as given in the question)
Simplifying the equation:
t' = (0 - 30) / (-0.5a)
t' = -30 / (-0.5a)
t' = 60 / a
We want to show that t' is six times as long as t:
t' = 6t
60 / a = 6 * (10 / a)
60 = 60
This equation is true, so it confirms that the time spent decelerating (t') is indeed six times as long as the time spent accelerating (t).
Now, let's find the total distance traveled by the car. The distance traveled during acceleration (d1) can be calculated using the equation:
d1 = ut + (1/2)at^2
Substituting the given values:
d1 = 20t + (1/2)a(t^2)
Since we know that t = 10 / a, we can substitute it into the equation:
d1 = 20 * (10 / a) + (1/2)a((10 / a)^2)
d1 = 200 / a + 100
The distance traveled during deceleration (d2) can be calculated using the equation:
d2 = vt' + (1/2)(-0.5a)(t'^2)
Substituting the given values:
d2 = 0 * (60/a) + (1/2)(-0.5a)((60/a)^2)
d2 = 9000 / a^2
The total distance traveled is the sum of d1 and d2:
d1 + d2 = 200 / a + 100 + 9000 / a^2
This total distance is given as 2900 m:
200 / a + 100 + 9000 / a^2 = 2900
Multiplying through by a^2:
200a + 100a^2 + 9000 = 2900a^2
Rearranging the equation:
2900a^2 - 200a - 100a^2 - 9000 = 0
2800a^2 - 200a - 9000 = 0
Dividing through by 100:
28a^2 - 2a - 90 = 0
Now we can use the quadratic formula to solve for 'a'. The quadratic formula is:
a = (-b ± sqrt(b^2 - 4ac)) / (2a)
where:
a = 28
b = -2
c = -90
Substituting the values:
a = (-(-2) ± sqrt((-2)^2 - 4(28)(-90))) / (2(28))
a = (2 ± sqrt(4 + 10080)) / 56
a = (2 ± sqrt(10084)) / 56
Simplifying the equation:
a ≈ (2 ± 100.42) / 56
Since acceleration cannot have a negative value, we take the positive square root:
a ≈ (2 + 100.42) / 56
a ≈ 102.42 / 56
a ≈ 1.83 m/s^2 (approximately)
The time for which the car is accelerating (t) can be found by substituting the value of 'a' into the earlier equation:
t = 10 / a
t ≈ 10 / 1.83
t ≈ 5.46 seconds (approximately)
Finally, to find the speed of the car 40 seconds before it comes to rest, we can calculate its speed during deceleration:
v' = u + (-0.5a)t'
Substituting the given values:
v' = 30 + (-0.5 * 1.83) * 40
v' ≈ 30 + (-0.915) * 40
v' ≈ 30 - 36.6
v' ≈ -6.6 m/s (approximate negative speed)
Note that the negative sign indicates that the car is moving in the opposite direction. So, the speed of the car 40 seconds before it comes to rest is approximately -6.6 m/s.
To solve this problem, let's break it down into smaller steps:
Step 1: Find the time for which the car is accelerating.
We can use the equation:
v = u + at
Given:
Initial speed, u = 20 m/s
Final speed, v = 30 m/s
Acceleration, a = 1 m/s^2
Using the equation, we can solve for time:
30 = 20 + 1t
t = (30 - 20) / 1
t = 10 seconds
So, the car is accelerating for 10 seconds.
Step 2: Find the value of a.
We can use the equation:
v = u + at
Given:
Initial speed, u = 20 m/s
Final speed, v = 30 m/s
Time, t = 10 seconds
Using the equation, we can solve for acceleration:
30 = 20 + a * 10
10a = 30 - 20
10a = 10
a = 1 m/s^2
So, the value of a is 1 m/s^2.
Step 3: Find the time spent decelerating.
To find the time spent decelerating, we need to calculate the deceleration and then use it to find the time using the equation:
Final velocity, v = 0 (since the car comes to rest)
Initial velocity, u = 30 m/s
Acceleration, a = -0.5a (negative because it is deceleration)
Using the equation, we can solve for time:
0 = 30 + (-0.5a) * t_d
-0.5a * t_d = -30
t_d = -30 / -0.5a
t_d = 60 / a
Step 4: Show that the time spent decelerating is six times as long as the time spent accelerating.
We know the time spent accelerating is 10 seconds (from step 1) and let's substitute the value of a (from step 2) into the equation from step 3:
t_d = 60 / a
t_d = 60 / 1
t_d = 60 seconds
So, the time spent decelerating is 60 seconds.
To show that the time spent decelerating is six times as long as the time spent accelerating, we can calculate the ratio:
t_d / t_a = 60 / 10 = 6
Therefore, the time spent decelerating is six times as long as the time spent accelerating.
Step 5: Find the speed of the car 40 seconds before it comes to rest.
To find the speed of the car at a given time, we can use the equation:
v = u + at
Given:
Initial speed, u = 30 m/s
Acceleration, a = -0.5a (since it is deceleration)
Time, t = 40 seconds
Using the equation, we can solve for velocity:
v = 30 + (-0.5a) * 40
v = 30 - 20
v = 10 m/s
So, the speed of the car 40 seconds before it comes to rest is 10 m/s.