We read your photo as:

A velocity-time graph for a car shows acceleration for the first 2s, constant velocity for the next 4s, and deceleration for the final 2s. The maximum velocity reached is 8 m/s. Find the total distanc

Physics

ScanSolve AI Answer

Answers

2 parts
a

48 m\text{48 m}

b

4 m/s2\text{4 m/s}^2

Step-by-step solution

2 steps

Here's the solution to question 8:

The figure shows a velocity-time graph for a car.
From the graph, we can identify the following:

  • Maximum velocity (VmV_m) = 8 m/s.
  • Time at which acceleration ends and constant velocity begins = 2 s.
  • Time at which constant velocity ends and deceleration begins = 6 s.
  • Total time for the journey = 8 s.

Part a· 1 step

Find the total distance travelled by the car.

The total distance travelled is the area under the velocity-time graph. This graph forms a trapezium.
The area of a trapezium is given by the formula:

Area=12×(sum of parallel sides)×height\text{Area} = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}

In this case:

  • The length of the top parallel side (duration of constant velocity) = 6 s−2 s=4 s6 \text{ s} - 2 \text{ s} = 4 \text{ s}.
  • The length of the bottom parallel side (total time) = 8 s8 \text{ s}.
  • The height of the trapezium (maximum velocity) = 8 m/s8 \text{ m/s}.

  1. Calculate the total distance.

    Total distance=12×(4 s+8 s)×8 m/s\text{Total distance} = \frac{1}{2} \times (4 \text{ s} + 8 \text{ s}) \times 8 \text{ m/s} Total distance=12×12 s×8 m/s\text{Total distance} = \frac{1}{2} \times 12 \text{ s} \times 8 \text{ m/s} Total distance=6 s×8 m/s\text{Total distance} = 6 \text{ s} \times 8 \text{ m/s} Total distance=48 m\text{Total distance} = 48 \text{ m}

    The total distance travelled by the car is 48 m\boxed{\text{48 m}}.

Part b

4 m/s2\text{4 m/s}^2

Calculate the deceleration of the car.

Deceleration occurs in the last phase of the journey, from t=6t=6 s to t=8t=8 s.
Deceleration is the negative of the acceleration, which is the gradient of the velocity-time graph during this period.

Deceleration=−(Change in velocityChange in time)\text{Deceleration} = - \left( \frac{\text{Change in velocity}}{\text{Change in time}} \right)
  • Initial velocity (uu) at t=6t=6 s = 8 m/s.
  • Final velocity (vv) at t=8t=8 s = 0 m/s.
  • Time interval (Δt\Delta t) = 8 s−6 s=2 s8 \text{ s} - 6 \text{ s} = 2 \text{ s}.

  1. Calculate the deceleration.

    Acceleration=v−uΔt\text{Acceleration} = \frac{v - u}{\Delta t} Acceleration=0 m/s−8 m/s2 s\text{Acceleration} = \frac{0 \text{ m/s} - 8 \text{ m/s}}{2 \text{ s}} Acceleration=−8 m/s2 s\text{Acceleration} = \frac{-8 \text{ m/s}}{2 \text{ s}} Acceleration=−4 m/s2\text{Acceleration} = -4 \text{ m/s}^2

    The deceleration is the magnitude of this negative acceleration.

    Deceleration=4 m/s2\text{Deceleration} = 4 \text{ m/s}^2

    The deceleration of the car is 4 m/s2\boxed{\text{4 m/s}^2}.

Was this helpful?

Try one yourself

A new problem that tests the same skill.

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question

More Physics Questions

QA car's velocity-time graph is shown. Find the total distance covered and the deceleration.QThe figure below shows velocity time graph for a car in metres. a) Find the total distance covered…QThe distance -time graph below shows the journey of a car. Convert the the distance -time graph to…QA car starts with a speed of 24 m/s. It decelerates for 3 seconds at 2 m/s2, then travels at a…QThe figure shows a velocity-time graph for a car.
Browse by topic:Mechanics
See all Physics questions →
Got a syllabus? Photograph it and I'll tell you what's due and what's worth the most.
Handwritten step-by-step solution preview

Handwritten Step-by-Step Solution

Get this as a handwritten image on WhatsApp — like a tutor wrote it for you.

Try on WhatsApp