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Montano1993 [528]
3 years ago
5

A car traveling 23 m/s begins to decelerate at a constant rate of 5 m/s^2 . After how many seconds does the car come to a stop?

(Use symbolic notation and fractions where needed.)
Physics
1 answer:
Tasya [4]3 years ago
7 0

Answer:

t = 4.6 s

Explanation:

given,

initial velocity of the car = 23 m/s

declaration rate of car = 5 m/s²

final velocity of the car = ?

time taken to stop the car = ?

using equation of motion to solve the question

v = u + a t

v is final velocity

u is the initial velocity

0 = 23 - 5 t

we have used negative sign because there is deceleration

5 t = 23

t = \dfrac{23}{5}

t = 4.6 s

time taken to stop the car is equal to t = 4.6 s

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jolli1 [7]

Answer:

Explanation:

Given that:

mass of stone (M) = 0.100 kg

mass of bullet (m) = 2.50 g = 2.5  ×10 ⁻³ kg

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Initial velocity of bullet (u_{bullet}) = (500 m/s)i

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Suppose we represent (v_{stone}) to be the velocity of the stone after the truck, then:

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m*u_{bullet} + M*{u_{stone}}= mv_{bullet}+Mv_{stone}

(2.50*10^{-3} \ kg) (500)i+0 = (2.50*10^{-3} \ kg)(300 \ m/s)j + (0.100 \ kg)v_{stone}

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