Physics, asked by bhartichandwani1234, 1 year ago

A Truck starting from rest moves with a uniform acceleration of 0.2 m /s2

for 2 minutes.

Find

(a) the speed acquired

(b) the distance travelled.​

Answers

Answered by Vamprixussa
87

⊕ANSWER⊕

Initial velocity = 0 m/s

a = 0.2 m/s

T = 2 minutes = 120 seconds

a) V = ?

v = u + at

v = 0 + 0.2(120)

v = 24 m

b) s = ?

s = ut + \frac{1}{2} at^{2}

  = 0 + \frac{1}{2}  * 0.2 * 120 * 120

  = 1440 m

Answered by ravilaccs
0

Answer:

The speed acquired  24 \ ms^-1 and the total distance traveled is 1440 \ m.

Explanation:

Given

The bus starting from rest moves with a uniform acceleration of 0.1 m s-2 for 2 minutes

Find out

Acceleration and distance traveled

Solution

(a) Acceleration

The bus starts from rest.

Therefore, the initial velocity (u) = 0 \ m/s

Acceleration(a) = 0.2\ ms^-2

Time = 2\ minutes = 120 s

Acceleration is given by the equation a=\frac{(v-u)}{t}

Therefore, terminal velocity (v) = (at)+u

= (0.2\ ms^-2 * 120s) + 0\ ms^1

= 24ms^-1 + 0 ms^-1\\\\=24 ms^-1

Therefore, terminal velocity(v) = 24\ m/s

(b) As per the third motion equation,

Sincea = 0.2 ms^-2\\ v = 24 ms^-1\\ u = 0 ms^-1 \\ t = 120sthe following value for s (distance) can be obtained.

Distance, s =\frac{(v^2- u^2)}{2a}

=\frac{(24^2 - 0^2)}{2(0.2)}

Therefore, s = 1440\ m.

The speeds acquired  24 \ ms^-1 and the total distance traveled is 1440 \ m.

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