A car start from rest, moves with uniform acceleration and attains a velocity of 72km/h in 1.5 s it moves with uniform speed for 20 s and is then brought to rest in 10 s under uniform retardation, then total distance travelled using velocity -time graph is?

I ALSO GO THIS GOT AS A QUESTION IN MY EXAM I GOT SOMEWHAT 400 ABOVE AS AN ANSWER

The area under the velocity-time graph is distance.

change 72km/hr to m/s. 72km/hr*1hr/3600s=20m/s

area=(20/2)m/s*1.5s +20m/s*20sec +(20/2)*10

check that...

Plz solve the above problem

To find the total distance traveled using the velocity-time graph, we first need to analyze the graph and divide it into different segments representing the motion of the car.

Let's break down the motion into three parts:

1. Part 1: Acceleration phase (0 s to 1.5 s)
During this phase, the car starts from rest and reaches a velocity of 72 km/h (which is equal to 20 m/s). From the given information, we know that the car moves with uniform acceleration.
To find the distance covered during this phase, we can use the equation:
v = u + at

Where:
v = final velocity (20 m/s),
u = initial velocity (0 m/s),
a = acceleration, and
t = time (1.5 s).

To find the acceleration, we can rearrange the formula as follows:
a = (v - u) / t

Substituting the values, we get:
a = (20 - 0) / 1.5
= 20 / 1.5
≈ 13.33 m/s^2

Now, we can use the equation for distance covered during uniform acceleration:
s = ut + (1/2)at^2

Substituting the values, we get:
s = 0 * 1.5 + (1/2) * 13.33 * (1.5)^2
= 0 + (1/2) * 13.33 * 2.25
≈ 15 m

Therefore, during the acceleration phase, the car covers approximately 15 meters.

2. Part 2: Uniform speed phase (1.5 s to 21.5 s)
During this phase, the car moves with a constant speed for 20 seconds. Since the velocity is constant, there is no acceleration or deceleration involved. The formula to calculate the distance covered during this phase is simply:
s = v * t

Where:
v = constant velocity (20 m/s), and
t = time duration (20 s).

Substituting the values, we get:
s = 20 * 20
= 400 m

Therefore, during the uniform speed phase, the car covers 400 meters.

3. Part 3: Retardation phase (21.5 s to 31.5 s)
During this phase, the car comes to rest over a duration of 10 seconds under uniform retardation (negative acceleration). This means the car slows down with a constant deceleration.
To find the distance covered during this phase, we can again use the equation:
s = ut + (1/2)at^2

Where:
u = initial velocity (20 m/s),
a = retardation, and
t = time (10 s).

Since the final velocity is 0 m/s, we can re-arrange the equation as:
0 = u + at

Substituting the values, we get:
0 = 20 + a * 10
=> a = -20/10
= -2 m/s^2

Now, we can use the equation to find the distance covered:
s = ut + (1/2)at^2

Substituting the values, we get:
s = 20 * 10 + (1/2) * (-2) * (10)^2
= 200 + (-1) * 100
= 200 - 100
= 100 m

Therefore, during the retardation phase, the car covers 100 meters.

Finally, adding up the distances covered in each phase:
Total distance = Distance during acceleration + Distance during uniform speed + Distance during retardation
= 15 m + 400 m + 100 m
= 515 m

Hence, the total distance traveled using the velocity-time graph is 515 meters.