Question

A car accelerates uniformly from rest and after 12 seconds has covered 40m. Calculate its acceleration and its final velocity.

Answer 1

Answer:

Explanation:s=ut+1/2×at²⇒40=(0)×12+1/2×a(12)²⇒a=40/72

v=u+at⇒v=(0)+40/72×12=40/6

Answer 2

The acceleration and final velocity of the car are $0.56m/s^2$ and 6.7 m/s respectively.

Explanation:

To get acceleration of the car use the second equation of motion as

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

Here, s is the distance covered by the car in time t and u is the initial velocity of the car and a is its acceleration.

Given s = 40 m, and t = 12 s and u = 0 (because car accelerates uniformly from rest).

Substitute the given values, we get

$40m=0+\frac{1}{2} a\times(12s)^2$

$a=\frac{80m}{144s^2}=0.56m/s^2$

Now to get final velocity use first equation of motion,

$v=u+at$

substitute the values, we get

$v=0+0.56m/s^2\times12s$

v = 6.7 m/s.

Thus, the acceleration and final velocity of the car are $0.56m/s^2$ and 6.7 m/s respectively.

#Learn More: equations of motion.

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