Question #ab83f

Question

Question #ab83f

1 Answer

Impact of this question

1567 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-24T08:21:25

Let

$East \to + ve direction of x-axis$ $"East"-> +"ve direction of x-axis"$

$West \to - ve direction of x-axis$ $"West"->-"ve direction of x-axis"$

$North \to + ve direction of y-axis$ $"North"-> +"ve direction of y-axis"$

$South \to - ve direction of y-axis$ $"South"->-"ve direction of y-axis"$

$\to V_{i} \to Initial velocity vetor$ $vec(V_i)->"Initial velocity vetor"$

$\to V_{f} \to Final velocity vetor$ $vec(V_f)->"Final velocity vetor"$

$∣ ∣ ∣ \to V_{i} ∣ ∣ ∣ = 8.5 \frac{m}{s}$ $abs(vec(V_i))=8.5m/s$

$∣ ∣ ∣ \to V_{f} ∣ ∣ ∣ = 19 \frac{m}{s}$ $abs(vec(V_f))=19m/s$

$t \to Time of change of velocity = 2 s$ $t->"Time of change of velocity"=2s$

$Direction of \to V_{i} = S 42^{\circ} W = {(270 - 42)}^{\circ}$ $"Direction of "vec(V_i)=S42^@W=(270-42)^@$

$Direction of \to V_{f} = W 35^{\circ} N = {(180 - 35)}^{\circ}$ $"Direction of "vec(V_f)=W35^@N=(180-35)^@$

$\to V_{i} = 8.5 (cos (270 - 42) ˆ i + sin (270 - 42) ˆ j)$ $vec(V_i)=8.5(cos(270-42)hati+sin(270-42)hatj)$
$= - 8.5 (sin 42 ˆ i + cos 42 ˆ j)$ $=-8.5(sin42hati+cos42hatj)$
$= - 5.69 ˆ i - 6.32 ˆ j \frac{m}{s}$ $=-5.69hati-6.32hatjm/s$

$\to V_{f} = 19 (cos (180 - 35) ˆ i + sin (180 - 35) ˆ j)$ $vec(V_f)=19(cos(180-35)hati+sin(180-35)hatj)$
$= 19 (- cos 35 ˆ i + sin 35 ˆ j)$ $=19(-cos35hati+sin35hatj)$
$= - 15, 56 ˆ i + 10.9 ˆ j \frac{m}{s}$ $=-15,56hati+10.9hatjm/s$

$Now average acceleration$ $"Now average acceleration"$
$\to a = \frac{\to V_{f} - \to V_{i}}{t}$ $vec(a)=(vec(V_f)-vec(V_i))/t$

$= \frac{1}{2} (- 9.87 ˆ i + 17.22 ˆ j) m s^{- 2}$ $=1/2(-9.87hati+17.22hatj)ms^-2$

$∣ ∣ \to a ∣ ∣ = \frac{1}{2} \sqrt{{9.87}^{2} + {17.22}^{2}} \approx 19.85 m s^{- 2}$ $abs(vec(a))=1/2sqrt(9.87^2+17.22^2)~~19.85ms^-2$

Science

Math

Humanities

... and beyond

Question #ab83f

1 Answer

Related questions

Impact of this question