An object have a velocity v-m-s at time t-0- It undergoes a constant acceleration a-2-ms-2 for 4 second- The magnitude of velocity at the end of 4 s is-

The correct option is A √(74) m/s Given, #10; #10;Initial velocity, u⃗=( î+ĵ #10;) ms^ 1 #10; #10;Acceleration, a⃗=( î 2ĵ #10;) ms ...

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Byju's Answer

Standard XII

Physics

Instantaneous Velocity

An object hav...

Question

An object have a velocity $\to v = (^i +^j) m / s$ at time $t = 0$ . It undergoes a constant acceleration $a = (^i - 2^j) m s^{- 2}$ for $4$ second. The magnitude of velocity at the end of $4 s$ is:

\sqrt{74} m / s

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\sqrt{80} m / s

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\sqrt{97} m / s

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\sqrt{47} m / s

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Solution

The correct option is A $\sqrt{74} m / s$
Given,

Initial velocity, $\to u = (^i +^j) m s^{- 1}$

Acceleration, $\to a = (^i - 2^j) m s^{- 2}$

Apply equation of kinematic

$\to v = \to u + \to a t$

$\to v = (^i +^j) + (^i - 2^j) \times 4$

$\to v = 5^i - 7^j$

$| \to v | = \sqrt{5^{2} + 7^{2}} = \sqrt{74} m s^{- 1}$

Final velocity of object is $\sqrt{74} m s^{- 1}$

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An object have a velocity →v=(^i+^j)m/s at time t=0. It undergoes a constant acceleration a=(^i−2^j)ms−2 for 4 second. The magnitude of velocity at the end of 4 s is:

The correct option is A √74 m/sGiven, Initial velocity, →u=(^i+^j)ms−1 Acceleration, →a=(^i−2^j)ms−2 Apply equation of kinematic →v=→u+→at →v=(^i+^j)+(^i−2^j)×4 →v=5^i−7^j |→v|=√52+72=√74ms−1 Final velocity of object is √74ms−1

An object have a velocity $\to v = (^i +^j) m / s$ at time $t = 0$ . It undergoes a constant acceleration $a = (^i - 2^j) m s^{- 2}$ for $4$ second. The magnitude of velocity at the end of $4 s$ is: