smc-4273-20-Velocity of objects

PHYSICS, M1 EQ-Bank 4 MC

A truck is driving along a straight road travelling at 10 ms\(^{-1}\). He then accelerates according to the acceleration time graph below:

Determine the final speed of the truck after it accelerates for 7 seconds.

\(20.5\ \text{ms}^{-1}\)
\(21.5\ \text{ms}^{-1}\)
\(22.5\ \text{ms}^{-1}\)
\(23.5\ \text{ms}^{-1}\)

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\(A\)

Show Worked Solution

The increase in velocity of the truck will be equal to the area under the acceleration time graph.
Speed Increase \(=\dfrac{1}{2} \times 7 \times 3 = 10.5\ \text{ms}^{-1}\)
Final speed \(=10.5 + 10 = 20.5\ \text{ms}^{-1}\)

\(\Rightarrow A\)

PHYSICS, M1 EQ-Bank 7

A plane is travelling at 315 ms\(^{-1}\) north when it passes through a dense cloud and slows down to a velocity of 265 ms\(^{-1}\) for safety precautions.

The plane did not change direction and travelled 2.5 km while it was slowing down.

Using north as the positive direction for all calculations, determine:

the change in velocity of the plane. (1 mark)

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the plane's acceleration. (2 marks)

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the time over which the plane slowed down. (2 marks)

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a. \(\text{50 ms}^{-1}\ \text{south}\)

b. \(\text{5.8 ms}^{-2}\ \text{south}\)

c. \(\text{8.62 s}\)

Show Worked Solution

a.	\(\Delta v\)	\(=v-u\)
		\(=265-315\)
		\(=-50\ \text{ms}^{-1}\)
		\(=50\ \text{ms}^{-1}\ \text{south}\)

b. Using \(v^2=u^2 +2as\) (time is not given):

\(a\)	\(=\dfrac{v^2-u^2}{2s}\)
	\(=\dfrac{(265)^2-(315)^2}{2 \times 2500}\)
	\(=-5.8\ \text{ms}^{-2}\)
	\(=5.8\ \text{ms}^{-2}\) to the south.

c. Using \(v=u+at\):

\(t\)	\(=\dfrac{v-u}{a}\)
	\(=\dfrac{265-315}{-5.8}\)
	\(=8.62\ \text{s}\)

PHYSICS, M1 EQ-Bank 4

A truck travelling in a straight line with a speed of 60 ms\(^{-1}\) slows down and comes to rest over a period of 20 seconds.

Calculate the change in velocity of the truck. (2 marks)

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Determine the average acceleration of the truck across the 20 seconds. (2 marks)

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a. \(-60\ \text{ms}^{-1}\)

b. \(-3\ \text{ms}^{-2}\)

Show Worked Solution

a.	\(\Delta v\)	\(=v-u\)
		\(=0-60\)
		\(=-60\ \text{ms}^{-1}\)

b.	\(a\)	\(=\dfrac{v-u}{t}\)
		\(=\dfrac{-60}{20}\)
		\(=-3\ \text{ms}^{-2}\)

PHYSICS, M1 2013 HSC 4 MC

Students performed an investigation to determine the initial velocity of a projectile.

Which row correctly identifies a hazard of this investigation and a related precaution?

	Hazard	Safety precaution
A.	flying projectile	wearing safety glasses
B.	range of projectile	measuring the range with a tape measure
C.	enclosed shoes	limiting the range of the projectile
D.	safety glasses	flying projectile

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\(A\)

Show Worked Solution

A hazard is a danger and the only danger in the answer options is the flying projectile.
Wearing safety glasses will protect students eyes from being hit from the projectile.

\(\Rightarrow A\)