a car starting from rest is accelerated at 2m/s(square) for 10s, then moves at constant velocity for 20s and then decelerates at 1m/s(square), finally stops. how far does it travel during its trip?
please help me! I need in solution for that problem
Stage 1, acceleration
a = 2 m/s^2
v = 0 + 2 * 10 = 20m/s
x = 0 + 0*10 + .5 (2)(10^2) = 100 m
Stage 2, starting with v = 20 and x = 100
a = 0
v = 20 + 0 * 20 = 20
x = 100 + 20(20) +.5*0*20^2 = 500
stage 3, starting with v = 20 and x = 500
a = -1
v = 20 -1(t) = 20 -t
v = 0 at end so t = 20 for deacceleration
x = 500 + 20(20) -.5(20)^2
= 500 + 400 - 200
= 700
For this problem:
Change in speed over change in time is acceleration...
A car travels on a straight, level road. Starting from rest, the car is going 20 m/s at the end of 2.0 sec.
(a) What is the acceleration of the car during the first 2.0 seconds of motion?
In 7.0 more seconds, the car is going 40 m/s.
(b) What is the car's acceleration for this period of time (during the 7 seconds)
The car then slows to 19 m/s in 2.0 seconds.
(c) What is the acceleration of the car during this period of time?
(d) What is the overall average acceleration of the car for the total time?
(a)80m
To find the total distance traveled by the car, we need to break down the motion into three parts: the acceleration phase, the constant velocity phase, and the deceleration phase.
1. Acceleration Phase:
The initial velocity (u) is 0 m/s (starting from rest), and the acceleration (a) is 2 m/s². The time (t) for this phase is 10 seconds.
We can use the equation: distance = u * t + (1/2) * a * t²
distance = (0 * 10) + (1/2) * 2 * (10²)
distance = 0 + 1/2 * 2 * 100
distance = 100 meters
2. Constant Velocity Phase:
During this phase, the car is moving at a constant velocity. The formula for distance traveled during constant velocity is speed * time.
Given that the velocity is constant for 20 seconds, and the acceleration is 0, the distance traveled during this phase is:
distance = velocity * time = (2 m/s² * 10 seconds) = 20 meters
3. Deceleration Phase:
Now, the car slows down with a deceleration of 1 m/s² until it stops at rest. The initial velocity (u) is the same as the final velocity (v) is 0 m/s, and we need to find the time (t) it takes to stop.
We can use the equation: v = u + a * t
0 = u + (1 * t)
t = - u / a = -0 / -1 = 0 seconds
So, during the deceleration phase, the car doesn't travel any additional distance.
Therefore, the total distance traveled by the car is the sum of the distances from each phase: 100 meters + 20 meters + 0 meters = 120 meters.
So, the car travels a distance of 120 meters during its trip.