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3 years ago

A car travels in a straight line for 5 h at a constant speed of 72 km/h. What is it’s acceleration

Physics

1 answer:

Luda [366]3 years ago

4 0

Answer:

$a \approx \: 0.001 \: m {s}^{ - 2}$

Explanation:

Given:

$initial \: velocity \: (u) = 0 \\ \\ Final \: Velocity \: (v) = 72 km /h \\ \\ Time \: (t) = 5 \: hours \\ \\ Acceleration \: (a) =? \\ \\ \because \: a = \frac{v - u}{t} \\ \\ \therefore \: a = \frac{72 - 0}{5} \\ \\ a = \frac{72}{5} \\ \\ a = 14.5 \: km {h}^{ - 2} \\ \\ a = \frac{14.5 \times 1000}{3600\times 3600} \: m {s}^{ - 2} \\ \\ a = 0.00111882716 \\ \\ a \approx \: 0.001 \: m {s}^{ - 2}$

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This question can be solved by using the equations of motion.

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b) It took the arrow "1.13 s" to reach a height of 17.5 m.

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where,

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Therefore,

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To find the time taken by the arrow to reach 17.5 m, we will use the second equation of motion again.

$h= v_it + \frac{1}{2}gt^2\\$

where,

g = acceleration due to gravity = 9.81 m/s²

h = height = 17.5 m

vi = initial speed = 9.81 m/s

t = time = ?

Therefore,

$17.5 = (9.81)t+\frac{1}{2}(9.81)t^2\\4.905t^2+9.81t-17.5=0$

solving this quadratic equation using the quadratic formula, we get:

t = -3.13 s (OR) t = 1.13 s

Since time can not have a negative value.

Therefore,

Learn more about equations of motion here:

brainly.com/question/20594939?referrer=searchResults

The attached picture shows the equations of motion in the horizontal and vertical directions.

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