Question

A car is traveling north at 17.7 m/s. After 12 s its velocity is 14.1 m/s in the same direction. Find the magnitude and direction of the car's average acceleration.

(A) 0.30 m/s2 , south
(B) 2.7 m/s2 , south
(C) 0.30 m/s2 , north
(D) 2.7 m/s2 , north

Answer 1

Answer:

$a)$ $0.3$ $m/s^{2}$ , $South$

Explanation:

Initial Velocity$(u)$ = $17.7$ $m/s$

Final Velocity$(v)$ = $14.1$ $m/s$

Time$(t)$ = $12$ $s$

Acceleration = $a$

Since They're all applied on the same axis we can solve this without using vectors.

we know that $v = u + at$

⇒$14.1 = 17.7 + 12a$

⇒$14.1-17.7=12a$

⇒$-3.6 = 12a$

⇒$-0.3 = a$

⇒Acceleration$(a) = -0.3$ $m/s^{2}$

Acceleration is negative, it means that it is applied in the opposite direction of velocity.

∴The Magnitude of Acceleration = $0.3$ $m/s^{2}$ The Direction is South.

⇒$a)$ $0.3$ $m/s^{2}$ , $South$

Answer 2

Answer:

Option A is the right choice.

Explanation:

As the car is traveling with a velocity of 17.7m/s towards the north and its final velocity after 12 seconds becomes 14.1m/s, it is retarding.

$V_{i} =17.7m/s\\V_{f}=14.1m/s \\t=12seconds$

The average acceleration of a body is given by a change in its velocity per unit of time.

$a=\frac{V_{2}- V_{1} }{T} \\a=\frac{14.1-17.7}{12}\\ a=0.30m/s^{2}$

Since the final velocity is lesser than the initial velocity body is retarding and acceleration is in the opposite direction of motion of the car.

Hence, the average acceleration of the body is 0.30m/s² in the south direction.