First slide

Instantaneous acceleration

Question

A point initially at rest moves along x-axis. Its acceleration varies with time as $a = (6 t + 5) m / s^{2} .$ If it starts from origin, the distance covered in 2s is:

Easy

A

20 m
B

18 m
C

16 m
D

25 m

Solution

$a = 6 t + 5 v = \int 6 t + 5 d t = 3 t^{2} + 5 t s = \int 3 t^{2} + 5 t d t = t^{3} + \frac{5}{2} t^{2} F r o m 0 t o 2 s, d i s \tan c e = 2^{3} + \frac{5}{2} \times 2^{2} = 18$

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