the velocity time graph of a car start from rest and it accelerate uniformly at a rate 3m/s2 for 5seonds if attain the velocity which maintain for 1m the car then brought to rest by a uniform retardation after another 30m.
draw the velocity time graph.
find the acceleration.
find the retardation.
total distance 1/2{AB+CE}h.
I wish it
To draw the velocity-time graph, we need to analyze the given information about the car's motion:
1. From the start to time t = 5s:
The car is accelerating uniformly at a rate of 3 m/s^2. At t = 0s, the car has a velocity of 0 m/s since it starts from rest. We can use the formula v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. Plugging in the values, we get v = 0 + (3 m/s^2) * (5s) = 15 m/s.
2. From t = 5s to t = 6s:
The car maintains a velocity of 15 m/s for 1 second. Since the velocity is constant, the graph will be a straight horizontal line at 15 m/s for this time interval.
3. From t = 6s to t = 12s:
The car is brought to rest by a uniform retardation. The distance covered in this time period is given as 30m. We can use the formula v^2 = u^2 + 2ad, where v is the final velocity, u is the initial velocity, a is the deceleration (retardation), and d is the distance. We will solve for a. Plugging in the values, we get 0^2 = (15 m/s)^2 + 2(a)(30m). Simplifying the equation, we get 0 = 225 m^2/s^2 + 60a m. Solving for a, we find a = -3.75 m/s^2.
Now, let's draw the velocity-time graph:
Time (s) | Velocity (m/s)
-------------------------
0 | 0
1 | 3
2 | 6
3 | 9
4 | 12
5 | 15
6 | 15
7 | 15
8 | 15
9 | 0
10 | 0
11 | 0
12 | 0
The acceleration during the first 5 seconds is 3 m/s^2.
The retardation (deceleration) after 6 seconds is -3.75 m/s^2.
To find the total distance covered, we'll consider the area under the velocity-time graph.
1/2{AB+CE}h = 1/2 * (5s * 15m/s + 6s * 15m/s) = 105m.