AP Calculus AB
Day 3
Section 4.1
7/29/2017
Perkins
The maker of a certain car advertises that it takes 12 seconds
to accelerate from 10 ft/sec to 40 ft/sec. Assuming constant
acceleration, compute
We know…
A t   k
a.
S 0  0
V  0  10 ft sec
the acceleration of the car.
V  t    k dt
 kt  C
V  t   52 t  10
V 0  k 0  C
10  C
V 10  40 ft sec
V 12  k 12  10
40  12k  10
5
k
2
The acceleration is
5 ft
2 sec 2
.
b. the distance the car travels in that 12 seconds.
Distance  S 12  S  0
Distance  S 12  0
Distance  300 feet
S  t   V  t  dt
S  t     52 t  10  dt
 54 t 2  10t  C
S 0 
5
4
0
2
 10  0   C
0 C
S  t   54 t 2  10t  0
S 12  
5
4
12
2
S 12  300 ft
 10 12 
AP Calculus AB
Day 3
Section 4.1
Perkins
2. The maker of a certain car advertises that it takes 12
seconds to accelerate from 10 ft/sec to 40 ft/sec. Assuming
constant acceleration, compute
a.
the acceleration of the car.
b.
the distance the car travels in that 12 seconds.
1. A ball is thrown upward with an initial velocity of V0 from a
height of S0 feet. Show that the position of the object at any
time t is given by S  t   16t 2  V0t  S0 .
We know…
A  t   32 ft sec2
V  0  V0
V  t    A(t ) dt   32dt  32t  C
S  0   S0
V  0  32  0   C
V0  C
V  t   32t  V0
S  t   V (t ) dt    32t  V0  dt  16t 2  V0t  C
S  0   16  0   V0  0   C
S0  C
2
S  t   16t 2  V0t  S0
1.
A ball is thrown upward with an initial velocity of V0 from
a height of S0 feet. Show that the position of the object at any
time t is given by S  t   16t 2  V0t  S0 .