(a) The velocity-time graph of a car is given below. The car weighs 1000 kg. (I) What is the distance travelled by the car in the first 2 seconds? (ii) What is the breaking force at the end of 5 seconds to bring the car to a stop within one second. (b) Derive the equation s=ut+1/2at square using graphical method.
Answers
velocity = distance / time
(1) distance = velocity * time = 15 * 2 = 30 m
(b)
distance = area of figure ABCFE
=area of triangle ABE + area of square BCFE
= 1/2 * at*t + ut
=>
s = ut + 1/2 at^2
(i) The distance travelled by the car is first 2 seconds is 15 m
(ii) The braking force at the end of 5 seconds to bring the car to a stop within one second is 15000 N
Explanation:
Given mass of car m = 1000 kg
(a) We know that in a velocity time graph, the distance travelled in time t = area under the velocity-time graph upto time t
Therefore, the distance travelled by the car in first 2 seconds
= Area of ΔABE
m
(b) From graph it is clear that the car moves with a constant acceleration from A to B and then with a zero acceleration from B to C and then with negative acceleration from C to D
Acceleration at the end of 5 seconds
= Acceleration (a) from C to D = Slope of line CD
=
= m/s²
Therefore the breaking force at the end of 5 seconds
N
(negative sign indicates that force applied is in the opposite direction of motion)
(b) Distance travelled from C to D
s = Area of ΔCFD
m
We have calculated, acceleration a from velocity C to velocity D as -15 m/s²
Initial velocity at C, u = 15 m/s
Time interval t = 6-5 = 1 seconds
Therefore,
Hope this answer is helpful.
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