Question

A material point that at instant t=0 is positioned at the origin is moving in a uniformly accelerated mode with acceleration a=-1,5 m/s^2. Find the expression for instantaneous velocity knowing that at t=4s the velocity is =15m/s. Also calculate the distance the material point covered in the first 5 s. Please use derivates and equations.

answer is: v=21-1,5t. and s= 86,3 m

Answer 1

Here we will use newtons 1st equation of motion

v=u+at

v final speed, u- initial speed,a- acclertion

Here at. t=4. v=15

15=u-1.5*4

u=21

so iinstantaneous velocity = 21-1.5t

According to 2nd equation of motion

s=ut+0.5at²

s - distance

For t=5

​s=21*5-0.5*1.5*5²​​​​​

s=​​​​​​ 86. 25 m