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A truck initially at rest, accelerates to a speed of 50km/h in 45 seconds. The truck remain at this speed for 5 minutes. The truck then brakes to a halt in 27 seconds. Draw a velocity time graph

1 determine the acceleration,
2 decerelation
3 the total distance travelled over the complete trip

User DanielC
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1 Answer

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Certainly! Let's break it down step by step:

Step 1: Drawing the velocity-time graph

The velocity-time graph will consist of three segments: acceleration, constant speed, and deceleration. On this graph, the y-axis will represent velocity in km/h, and the x-axis will represent time in seconds.

- From 0 to 45 seconds, the truck accelerates, so the graph will show a line sloping upwards from 0 to 50 km/h.

- From 45 seconds to 5 minutes and 45 seconds (which is 345 seconds), the truck maintains a constant speed of 50 km/h. So, the graph will have a horizontal line at 50 km/h.

- From 345 to 372 seconds, the truck decelerates to a stop, so the graph will show a line sloping downwards back to 0 km/h.

Step 2: Determine the acceleration

Acceleration is the change in velocity divided by the time taken. The truck goes from 0 to 50 km/h (13.9 m/s, since 50 km/h is equal to 50*1000/3600 meters per second) in 45 seconds.

Acceleration (a) = change in velocity / time taken

a = (50 km/h - 0) / 45 seconds

= 13.9 m/s / 45 s

≈ 0.31 m/s²

Step 3: Determine the deceleration

Deceleration is the decrease in velocity divided by the time taken. The truck goes from 50 km/h to 0 in 27 seconds.

Deceleration (d) = change in velocity / time taken

d = (0 - 50 km/h) / 27 seconds

= -13.9 m/s / 27 s

≈ -0.51 m/s²

Note that deceleration is negative as it is in the opposite direction of the velocity.

Step 4: Determine the total distance traveled

The total distance traveled can be calculated by finding the area under the velocity-time graph.

1. For the acceleration part: the area is a triangle with base 45 s and height 50 km/h.

Area1 = (1/2) * base * height

= (1/2) * 45 s * 13.9 m/s

≈ 312.75 meters

2. For the constant speed part: the area is a rectangle with length 300 s (5 minutes) and height 50 km/h.

Area2 = length * height

= 300 s * 13.9 m/s

≈ 4170 meters

3. For the deceleration part: the area is again a triangle with base 27 s and height 50 km/h.

Area3 = (1/2) * base * height

= (1/2) * 27 s * 13.9 m/s

≈ 187.65 meters

Total distance = Area1 + Area2 + Area3

≈ 312.75 + 4170 + 187.65

≈ 4670.4 meters (or about 4.67 kilometers).

So, the truck accelerates with an approximate acceleration of 0.31 m/s², decelerates with about -0.51 m/s², and covers a total distance of about 4.67 kilometers during its trip.

User Jhavatar
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