Question

A car staring from rest accelerates at 3.0 m/s2

for 8.0 secs. Calculate the distance

travelled by the car​

Answer 1

Answer:

A car starting from rest accelerates at a constant rate of 3.0 m/s2 for 8.0 seconds. How far does the car travel during this time?

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The equation for problems of this sort is as follows:

distance = (1/2)(a)(t^2)

distance = (1/2)(3.0 m/sec^2) x (8.0 sec^2)

distance = 96 meters

You can also think about it this way. With constant acceleration, the average velocity will occur at the midpoint of the time or (1/2)(8.0 sec)(3.0 m/sec^2) or 12 m/sec.

Then 12 m/sec x 8.0 sec = 96 meters

Answer 2

$u \: = 0 \\ \\ a = 3 \: ms {}^{ - 2} \\ \\ t = 8 \: sec \\ \\ s \: = ? \\ \\ = > \: s = ut + \frac{1}{2} at {}^{2} \\ \\ = 0 \times 8 + \frac{1}{2} \times 3 \times 8 {}^{2} \\ \\ = 0 + \frac{1}{2} \times 3 \times 64 \\ \\ = \frac{1}{2} \times 3 \times 64 \\ \\ = 3 \times 32 \\ \\ = 96 \: m$

I hope this helps.