Question

A particle is moving with a velocity of 60.0 m/s in the positive x direction at t= 0. Between t= 0 and t= 15.0 s the velocity decreases uniformly to zero. What was the acceleration during this 15.0-s interval? What is the significance of the sign of your answer?

Answer 1

Answer:

Acceleration of the object is given as

$a = - 4 m/s^2$

so here negative sign shows that the direction of acceleration is opposite to the velocity of the object

Explanation:

As we know that velocity decreases uniformly from maximum to zero

so here in this uniform deceleration we know that

$v_i = 60 m/s$

$v_f = 0$

$t = 15 s$

now by using the equation of kinematics we have

$v_f - v_i = at$

$0 - 60 = a(15)$

$a = - 4 m/s^2$

so here negative sign shows that the direction of acceleration is opposite to the velocity of the object