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Annette [7]
3 years ago
14

A car starts from rest and travels for 5.0 s with a constant acceleration of -1.5 m/s^2. What is the final velocity of the car?

Physics
2 answers:
bulgar [2K]3 years ago
7 0

Answer:

Final velocity

= - 7.5 \frac{m}{sec}

Distance covered

= 18.75 m.

Explanation:

If an object having initial velocity u is applied an acceleration of a for t time, it's final Velocity v is given by v = u + at

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

We have u=o (as the object starts from rest),

t=5 sec and acceleration a

=  - 1.5   \frac{m}{ {sec}^{2} }

Final velocity v=

=  - 1.5 \times 5 =  - 7.5 \frac{m}{sec}

Therefore, The final velocity is :-

- 7.5 \frac{m}{sec}

________________________________

Extra Info(Distance Covered)

Distance covered-

S=\frac{1}{2} × (-1.5)× {5}^{2} = -18.75

= - 18.75

Hence, Distance Covered is - 18.75 m.

________________________________

Note-Minus sign indicates that acceleration is in reverse direction.

Alecsey [184]3 years ago
4 0
Hence final velocity v=−1.5×5=−7.5 msec . Distance covered S=12×(−1.5)×52=−18.75 m
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