Question

A car is traveling due north at 23.6 m>s. Find the velocity of the car after 7.10 s if its acceleration is (a) 1.30 m>s2 due north, or (b) 1.15 m>s2 due south.

Answer 1

Answer:

a) v = 32.8 m/s

b) v= 15.4 m/s

Explanation:

a)

• Applying the definition of acceleration (assumed to be constant), we can write the following expression for the velocity v:

        $v = v_{o} + a*t (1)$

        where v₀ is the initial velocity and a is the acceleration, being t the

       time elapsed.

• In the case a) the acceleration and the velocity vectors have both the same direction (due north), so both have the same sign, which means that the car is speeding up.
• Replacing by the givens v₀ and t in (1), we get:

       $v = 23.6 m/s + (1.30 m/s2 * 7.10 s) = 32. 8 m/s (2)$

b)  

• In this case, the acceleration vector and the velocity vector have opposite directions, so the car slows down, due to both vectors have opposite signs.
• Replacing by the givens in (1) and taking into account the signs, we get:

       $v = 23.6 m/s + ((-1.15 m/s2) * 7.10 s)) = 15. 4 m/s (3)$