Question

the velocity time graph of a truck is plotted below. (a) calculate the magnitude of displacement of truck in 15 seconds
(b)during which part of the journey was the truck decelerating
(c)calculate the magnitude of average velocity of the truck

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Answer 1

Given:

Velocity-Time graph of a truck has been plotted below.

To find:

• Magnitude of Displacement in 15 sec.

• Part of the graph where the truck is decelerating.

• Average Velocity of truck.

Calculation:

Displacement of any object is given by the area of its velocity-time graph.

Let Displacement be d;

$\sf{ \therefore \: d = area}$

$\sf{ = > \: d = \dfrac{1}{2}(4)(5) + (4)(12 - 5) + \dfrac{1}{2} (4)(15 - 12)}$

$\sf{ = > \: d = \dfrac{1}{2}(4)(5) + (4)(7) + \dfrac{1}{2} (4)(3)}$

$\sf{ = > \: d = 10+ 28 + 6}$

$\red{ \boxed {\sf{ = > \: d = 44 \: m}}}$

The truck is decelerating at the time interval of $\boxed{\rm{t = 12 \:sec \: to \:t = 15\: sec}}$.

Average Velocityis defined as the ratio of total displacement to the total time taken;

$\rm{avg. \: v = \dfrac{total \: displacement}{total \: time} }$

$\rm{ = > avg. \: v = \dfrac{44}{15} }$

$\blue{ \boxed{ \rm{ = > avg. \: v = 2.93 \: m {s}^{ - 1} }}}$

Answer 2

Explanation:

of its velocity-time graph.

Let Displacement be d;

\sf{ \therefore \: d = area}∴d=area

\sf{ = > \: d = \dfrac{1}{2}(4)(5) + (4)(12 - 5) + \dfrac{1}{2} (4)(15 - 12)}=>d=

2

1

(4)(5)+(4)(12−5)+

2

1

(4)(15−12)

\sf{ = > \: d = \dfrac{1}{2}(4)(5) + (4)(7) + \dfrac{1}{2} (4)(3)}=>d=

2

1

(4)(5)+(4)(7)+

2

1

(4)(3)

\sf{ = > \: d = 10+ 28 + 6}=>d=10+28+6

\red{ \boxed {\sf{ = > \: d = 44 \: m}}}

=>d=44m

The truck is decelerating at the time interval of \boxed{\rm{t = 12 \:sec \: to \:t = 15\: sec}}

t=12sectot=15sec

.

Average Velocityis defined as the ratio of total displacement to the total time taken;

\rm{avg. \: v = \dfrac{total \: displacement}{total \: time} }avg.v=

totaltime

totaldisplacement

\rm{ = > avg. \: v = \dfrac{44}{15} }=>avg.v=

15

44