一道物理英文题The speed of a bullet as it travels down the barrel of a rifle toward the opening is given by v = (–5.00 × 107)t2 + (3.00 × 105)t,where v is in meters per second and t is in seconds.The acceleration of the bullet just as it leaves the barrel is zero.(a) Determine the acceleration and position of the bullet as a function of time when the bullet is in the barrel.(b) Determine the length of time the bullet is accelerated.(c) Find the speed at whic
一道物理英文题
The speed of a bullet as it travels down the barrel of a rifle toward the opening is given by v = (–5.00 × 107)t2 + (3.00 × 105)t,where v is in meters per second and t is in seconds.The acceleration of the bullet just as it leaves the barrel is zero.
(a) Determine the acceleration and position of the bullet as a function of time when the bullet is in the barrel.
(b) Determine the length of time the bullet is accelerated.
(c) Find the speed at which the bullet leaves the barrel.
(d) What is the length of the barrel?
1、加速度a=dv/dt=2*（-5.00*10^7)t+3.00*10^5=-1.00*10^8t+3.00*10^5
位移S=∫vdt=1/3*(-5.00*10^7)t^3+1/2*(3.00*10^5)t^2=-1.67*10^7t^3+1.50*10^5t^2
2、a=0
-1.00*10^8t+3.00*10^5=0
t=3.00*10^-3秒=3.00毫秒
子弹加速的时间为3毫秒
3、子弹出口速度v=-5.00*10^7t^2+3.00*10^5t==-5.00*10^7*(3.00*10^-3)^2+3.00*10^5*3.00*10^-3=450米/秒
4、枪膛长度S=-1.67*10^7t^3+1.50*10^5t^2
=-1.67*10^7(3.00*10^-3)^3+1.50*10^5*(3.00*10^-3)^2
=0.90米
(a)a=dv/dt=2×（–5.00 × 107）t+(3.00 × 105) s=（1/3）×(–5.00 × 107)t3 + （1/2）×(3.00 × 105)t2(b)let a=0,then t=0.294393s(c)let t=0.294393,then v=46.3668m/s(d)let t=0.294393,then s=9.10003m...