Question

The position of a particle moving in a straight line is given by x = 3 + 4t + 3t2, where x is in metre and time is in second. Find the values of the following physical quantities for the particle at t = 2s. (i) (ii) (iii) (iv) 2Position Displacement Velocity Acceleration

Answer 1

Soln:

Given, X = 3t² + 4t + 3 ---------------(A)

(i) Displacement at t = 2 sec
 
 X = 3(2)² + 4(2) + 3 = 23 meter.

(ii) Velocity: Differentiate (A) with respect to time

V(t) = 6t + 4

velocity at t = 2 sec
V(2) = 6(2) + 4 = 16 m/s

(iii) Acceleration: Differentiate Velocity V(t) with resect to time.

a = 6 m/s²

Hope it helps!

Answer 2

Answer:

21m

Explanation:

v=dsdt=4t3+3t2−1

∫s0ds=∫21(4t3+3t2−1)dt

s=∣∣t4+t3−t∣∣21

={(2)4+(2)3−(2)}−{(1)4+(1)3−(1)}

=22−1=21m