brakes are applied to a moving truck to produce retardation of 5m+s2. if the time taken between application of brakes and the truck to stop is 2s, calculate the distance travelled by the truck during this time.
The distance covered by the truck will be 10 m as,
v = final velocity = 0 ms-1 as it comes to rest
u = initial velocity = ?
t = time = 2 sec
a = acceleration = -5 ms-2 (it is negative as it is in the opposite direction)
s = distance = ? which is what we have to find
Using the equations of motion,
v = u + at
0 = u + (-5)(2)
0 = u + (-10)
0 = u - 10
Bringing 10 to the LHS
0 + 10 = u
u = 10 ms-1
By the second equation of motion
s = ut + 1/2at2
s = (10)(2) + 1/2(-5)(2)2
s = 20 + 1/2(-5)(2)(2)
s = 20 + (-5)(2)
s = 20 + (-10)
s = 20 - 10
s = 10 m
v = final velocity = 0 ms-1 as it comes to rest
u = initial velocity = ?
t = time = 2 sec
a = acceleration = -5 ms-2 (it is negative as it is in the opposite direction)
s = distance = ? which is what we have to find
Using the equations of motion,
v = u + at
0 = u + (-5)(2)
0 = u + (-10)
0 = u - 10
Bringing 10 to the LHS
0 + 10 = u
u = 10 ms-1
By the second equation of motion
s = ut + 1/2at2
s = (10)(2) + 1/2(-5)(2)2
s = 20 + 1/2(-5)(2)(2)
s = 20 + (-5)(2)
s = 20 + (-10)
s = 20 - 10
s = 10 m