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A car of mass $1250{\text{ }}kg$ travels along a horizontal straight road. The power of the car’s engine is constant and equal to $24{\text{ }}kW$ and the resistance to the car’s motion is constant and equal to $R{\text{ }}N$. The car passes through the point $A$ on the road with speed $20{\text{ }}m{\text{ }}{s^{ - 1}}$ and acceleration $0.32{\text{ }}m{\text{ }}{s^{ - 2}}$.

a) Find the value of $R$.

The car continues with increasing speed, passing through the point $B$ on the road with speed $29.9{\text{ }}m{\text{ }}{s^{ - 1}}$.

The car subsequently passes through the point $C$.

b) Find the acceleration of the car at $B$, giving the answer in $m{\text{ }}{s^{ - 2}}$ correct to 3 decimal places.

c) Show that, while the car’s speed is increasing, it cannot reach $30{\text{ }}m{\text{ }}{s^{ - 1}}$.

d) Explain why the speed of the car is approximately constant between $B$ and $C$.

e) State a value of the approximately constant speed, and the maximum possible error in this value at any point between $B$ and $C$.

The work done by the car’s engine during the motion from $B$ to $C$ is $1200{\text{ }}kJ$.

f) By assuming the speed of the car is constant from $B$ to $C$, find, in either order,

(i) the approximate time taken for the car to travel from $B$ to $C$,

(ii) an approximation for the distance $BC$.

پاسخ تشریحی :
نمایش پاسخ

a) $DF = 24000/20$

$\left[ {DF - R = 1250 \times 0.32} \right]$

$R = 800$

b) $24000/29.9 - 800 = 1250a$

Acceleration is $0.002m{s^{ - 2}}$

c) [$a = \left( {24000/30 - 800} \right)/1250$

$24000/v - 800 \gt 0 \to v \lt 30$]

Car not accelerating when $v = 30$ or

Speed cannot reach $30m{s^{ - 1}}$

d) $29.9 \leqslant v \lt 30 \to $  speed approximately constant

e) $30m{s^{ - 1}}$ (max error 0.1) or $29.95m{s^{ - 1}}$ (max error 0.05) or $29.9m{s^{ - 1}}$ (max error 0.1)

f)(i) $\left[ {24 = 1200/T} \right]$

Time taken is 50s

(ii) [${s = 30 \times 50}$ or ${29.95 \times 50}$ or ${29.9 \times 50}$]

Distance $BC$ is $1500m$ or $1500m$ or $1495m$

ALTERNATIVE FOR PART (f)

(ii) $\left[ {1200000 = 800d} \right]$

Distance $BC$ is $1500m$

(i) [$t = 1500/30$ or $1500/29.95$ or $1500/29.9$]

Time taken is 50s or 50.1s or 50.2s

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