Question

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?

Answer 1

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 :-

$- 7.5 \frac{m}{sec}$

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Extra Info(Distance Covered)

Distance covered-

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

$= - 18.75$

Hence, Distance Covered is $- 18.75 m.$

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

Answer 2

Hence final velocity v=−1.5×5=−7.5 msec . Distance covered S=12×(−1.5)×52=−18.75 m